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Unformatted text preview: Chapter Eleven INTERMOLECULAR FORCES AND LIQUIDS AND SOLIDS • • • • • The Kinetic Molecular Theory of Liquids and Solids Intermolecular Forces Properties of Liquids The Solid State Phase Changes and Phase Diagrams THE KINETIC MOLECULAR THEORY OF LIQUIDS AND SOLIDS STUDY OBJECTIVE 1. Describe the characteristic properties of solids, liquids, and gases. The Molecular View of Liquids and Solids. In this chapter we will examine the structure of liquids and solids, and discuss some of the fundamental properties of these two states of matter. Matter can exist in three states, or phases: gas, liquid, and solid. Each of these states will be described from a molecular viewpoint in terms of four characteristics: distance between molecules, attractive forces between molecules, the motion of molecules, and the orderliness of the arrangement of molecules. Figure 1.7 (text) gives a molecular representation of a solid, a liquid, and a gas. In Chapter 5 you learned that a gas can be pictured as a collection of molecules that are far apart, are in constant random motion, and exert almost no attractive forces on each other. The kinetic molecular theory was applied to explain the behavior of gases. Liquids and solids are quite different from gases. They are referred to as the condensed states because their molecules are very close together. Gases have low density, high compressibility, and completely fill a container. The condensed states have relatively high density, are almost incompressible, and have definite volumes. These properties indicate that the "molecules" of solids and liquids are close together and are held by strong intermolecular forces. One difference between solids and liquids is that liquids are fluids, but solids are not. In liquids, molecules can move past each other even though they cannot get very far from each other. In solids, molecules are fixed in position, and at most can only vibrate about that position. Many solids are characterized by long-range order of their constituent "molecules." This order gives rise to crystal structures. Table 11.1 summarizes some of the properties of the three states of matter. 2 11 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 12 / Intermolecular Forces and Liquids and Solids Table 11.1 Characteristic Properties of the States of Matter ____________________________________________________________________ Gas Liquid Solid ____________________________________________________________________ Assumes the volume Has a definite volume Has a definite volume and shape of its but assumes the shape and shape container of its container Is a fluid (flows readily) Is a fluid Is not a fluid Very compressible Only slightly compressible Virtually incompressible Low density High density High density Molecules far apart Molecules close together Molecules close together ____________________________________________________________________ INTERMOLECULAR FORCES STUDY OBJECTIVES 1. 2. Describe the different types of intermolecular attractive forces. Relate the strength of dispersion forces to molecular mass. van der Waals Forces. The forces that hold atoms together such as covalent bonds exist within molecules. The forces that hold individual molecules close together as in a solid or a liquid are called intermolecular forces (IMF). Intermolecular forces can be grouped for convenience into van der Waals forces, ion-dipole forces, and hydrogen bonds. The van der Waals forces cause a gas to deviate from ideal gas behavior and are also responsible for gases condensing to form liquids. They are attractive forces resulting from dipole-dipole forces, dipole-induced dipole forces, and dispersion forces. We shall discuss each of these interactions briefly. Dipole-Dipole Forces. In a dipolar substance, molecules tend to become oriented with the positive end of one molecule directed toward the negative ends of neighboring dipoles. The attraction is essentially electrostatic. The larger the dipole moment the larger the force. D ipole-dipole forces are attractive forces between polar molecules. Dipole-Induced Dipole Forces. The presence of a polar molecule in the vicinity of another molecule (usually nonpolar) has the effect of polarizing the second molecule. This means that a dipole moment has been induced in the previously nonpolar molecule. The induced dipole can then interact with the dipole moment of the first molecule, and the two molecules are attracted to each other. The dipole in the nonpolar molecule is called an i nduced dipole because it is due to the close proximity of a dipole or in some cases an ion. Dispersion Forces. Nonpolar molecules such as O 2 , SF 6 , and even the noble gases He, Ne, Ar, Kr, and Xe all show deviations from the ideal gas equation, and all condense at low temperatures to form liquids. We can ask: What kind of attractive forces exist between nonpolar molecules? In 1930, Fritz London proposed that although these molecules have no permanent dipole moments, their electron clouds are fluctuating. In a helium atom, for instance, the electrons occupy a 1s orbital, which has a spherical shape. The electrons are in constant motion, and if for an instant they should both move to the same side of the nucleus, a short-lived dipole will exist. In an instant the electron will continue its motion and the dipole will be gone, but a new one will be Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Intermolecular Forces and Liquids and Solids / 2 13 formed. These short-lived dipoles are called instantaneous dipoles or temporary dipoles. An instantaneous dipole can polarize a neighboring molecule producing an induced dipole. The two dipoles will tend to stick together. The attractive interactions caused by instantaneous dipoles are known as dispersion forces. The strength of dispersion forces depends on the polarizability of the molecule and can be as large as, or larger than, dipole-dipole forces. P olarizability is the tendency of an electron cloud to be distorted by the presence of an electrical charge such as that of an ion or the partial charge of a dipole. In general, the polarizability increases as the total number of electrons in a molecule increases. Since molecular mass and number of electrons are related, the polarizability of molecules and the strength of dispersion forces increases with increasing molecular mass. Ion-Dipole Forces. The electrostatic attraction between an ion and a polar molecule is called an i ondipole force. Hydration of ions, which was discussed in Section 6.6, is a good example of ion-dipole interactions. When an ionic compound such as NaBr dissolves its cations and anions are attracted to water molecules. Water molecules being polar have a negative end and a positive end. The Na+ ions attract the negative end of the water molecule, and the Br– ions attract the positive end of the water molecule. As the charge of an ion increases, it attracts polar molecules more strongly. Thus Mg2+ ions attract water molecules more strongly than Na+ ions, and S2– ions attract water molecules more strongly than Br– ions. Hydrogen Bonding. An additional type of intermolecular force is necessary to explain certain properties of ammonia, water, and hydrogen fluoride. The hydrogen bond is a special type of dipole-dipole interaction between the hydrogen in a polar bond and an O, N, or F atom. In water, for instance, the attractions are more than just the attractions of one dipole for another. Each hydrogen atom with its partial positive charge is attracted to one of the lone electron pairs of an oxygen atom of a neighboring molecule. This interaction is usually represented as follows: H δ+ | δ– H—O : · ·· H—O : | ˙˙ H .. and is called a hydrogen bond. Hydrogen bonding is limited to compounds containing nitrogen, oxygen, and fluorine. T he requirements for hydrogen bond formation are: 1. The element that is covalently linked to hydrogen must be sufficiently electronegative to attract bonding electrons and leave the hydrogen atom with a significant δ+ charge. 2. The electronegative atom, bound to hydrogen by the hydrogen bond, must have a lone pair of electrons. 3. The small size of the hydrogen atom allows it to approach nitrogen, oxygen, and fluorine atoms in neighboring molecules very closely. It is significant that hydrogen bonding is limited to these three elements of the second period. Both sulfur and chlorine are highly electronegative, but do not form hydrogen bonds. These third period atoms are apparently too large and do not present a highly localized nonbonding electron pair for the H atom to be attracted to. Hydrogen bonds are the strongest of the intermolecular forces, with energies of the order 10–40 kJ/mol. Van der Waals interactions correspond to between 2 and 20 kJ/mol. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 14 / Intermolecular Forces and Liquids and Solids _______________________________________________________________________________ EXAMPLE 11.1 Types of Intermolecular Forces Indicate all the different types of intermolecular forces that exist in each of the following substances. a. CCl 4 (l) b. HBr(l) c. CH3 OH(l) •Method of Solution First you must use the concepts of bond polarity and molecular structure to decide whether the molecule is polar or not. a. CCl 4 is nonpolar. The only type of IMF are dispersion forces. b. HBr is a polar molecule. The types of IMF are dipole-dipole and dispersion forces. There is no hydrogen bonding in HBr. The Br atom is not electronegative enough. c. CH3 OH is polar and has a hydrogen atom bound to an oxygen atom. Dipole-dipole attractions, dispersion forces, and H bonds exist in liquid methanol. _______________________________________________________________________________ EXAMPLE 11.2 Types of Intermolecular Forces Which of the following substances should have the strongest intermolecular attractive forces: N2 , Ar, F2 , or Cl2 ? •Method of Solution Note that none of these molecules is polar and that there is no chance for H bonding to occur. The only intermolecular forces existing in these molecules then are dispersion forces. Dispersion forces increase as the polarizability of the molecule increases, while polarizability increases with molecular mass. Answer: The most polarizable molecule will be Cl 2 (70.9 amu). _______________________________________________________________________________ EXAMPLE 11.3 Intermolecular Forces The dipole moment µ in HCl is 1.03 D, and in HCN it is 2.99 D. Which one should have the higher boiling point? •Method of Solution The larger the dipole moment, the stronger the IMF. The stronger the IMF, the higher the temperature needed to provide molecules of the liquid with enough kinetic energy to overcome these attractive forces. The boiling point of HCN should be greater than that of HCl. Observed values are b.p. (HCl) = –85°C, and b.p. (HCN) = 26°C. _______________________________________________________________________________ EXERCISES 1. 2. Back Indicate the types of intermolecular forces that exist between molecules (or basic units) in each of the following. a. PCl3 b . CO 2 c . Cl 2 d. ICl e. KCl Which of the following should have the strongest intermolecular forces? CH4 C l 2 C O CS 2 Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Intermolecular Forces and Liquids and Solids / 2 15 3. Which of the folowing species are capable of hydrogen bonding? a. CH 3 F b. HF c. CH 3 CH2 OH d. CH3 NH2 e . CH 4 4. Which member of each pair should have the higher boiling point? a. O 2 or CO b. Br2 or ICl c. H2 O or H2 S d. PH3 or AsH 3 5. What type of intermolecular forces must be overcome in order to: a. vaporize water b. dissociate H2 into H atoms c. boil liquid O2 PROPERTIES OF LIQUIDS STUDY OBJECTIVES 1. 2. Describe surface tension and viscosity. List six unusual properties of water. Surface Tension. The surface tension of a liquid is a property that has no direct counterpart in solids or gases. Surface tension is what makes water bead up on a freshly waxed surface and what makes soap bubbles round. Surface tension is a force that tends to minimize the surface area of a drop of liquid. Energy is required to expand the surface of a liquid, and the s urface tension i s the amount of energy required to increase the surface area of a liquid by a unit area. Liquids in which strong intermolecular forces exist exhibit high surface tensions. Water beads up on a freshly waxed surface because its cohesive forces are stronger than its adhesive forces. Cohesion is the intermolecular attraction between like molecules in the drop. A dhesion is the intermolecular attractions between unlike molecules, such as between water and wax. Since there is very little attraction between water and wax (adhesion), the strong cohesive forces tend to maintain the drop. The drop adopts spherical shape because a sphere has the least surface area. When water comes in contact with glass, adhesion is stronger than cohesion. Water is pulled against the glass surface, and we say that water "wets" the surface. Wetting results in the spreading of a thin film of water on the glass surface. The strength of the cohesive forces overcomes the ability of the surface tension to make a spherical drop. Viscosity. One characteristic of liquids that we have all observed is related to how freely they flow. Water pours much more freely than motor oil, and motor oil more readily than glycerol, for instance. The unique pouring characteristics of each liquid are the result of its viscosity. V iscosity is a measure of a fluids resistance to flow. Liquids whose molecules have strong intermolecular forces have greater viscosities than liquids that have weaker intermolecular forces. Table 11.3 (in the text) lists viscosities of some common liquids. The viscosity of water and glycerol is the result of strong hydrogen bonds. The viscosity of motor oil results from strong London dispersion forces. Properties of Water. Given its abundance and familiarity, we often fail to realize just how unusual water actually is. Some of the unusual properties of water are: 1. 2. 3. Back Water has a considerably greater surface tension than most other liquids. At 4.184 J/g·°C, the specific heat of water is one of the highest of all substances. The heats of fusion and vaporization are 6.0 and 40.8 kJ/mol, respectively. H 2 O is unusual because 40.8 kJ is a huge vaporization energy. Compare values in Table 11.6 (text). Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 16 / Intermolecular Forces and Liquids and Solids 4. Boiling point. The boiling point of water is about 200°C higher than might be reasonably expected. Boiling point tends to be related to the molar mass. Compare the boiling point of water with that of other low molar mass liquids in Table 11.2. Table 11.2 Boiling Points of Some Small Molecules _______________________________________ Molar Mass Boiling Pt Liquid (g/mol) (°C) _____________________________________ H2 O 18 100° CH4 16 –159° NH3 17 –33° _____________________________________ 5. Density versus temperature. As liquids cool they become more and more dense. As water is cooled from 100°C down to 4°C it does indeed get more dense. But from 4°C (actually 3.98°C) down to 0°C just the opposite happens: it gets less dense (see Table 11.3). Table 11.3 Density of H 2 O Near 0°C ___________________________________ Water Density Temperature °C (g/cm3 ) ___________________________________ 10.0 0.9997 5.0 0.99999 3.98 1.00000 2.0 0.99997 0 0.9998 ___________________________________ 6. Density of liquid versus density of solid. There are over seven million known chemical compounds. All, except a dozen or so, have a solid state that is more dense than the liquid. Water, of course, is one of the exceptions. The density of solid water at 0°C is 0.917 g/cm3 , which compares with 0.9998 g/cm3 for the liquid. Because of its lower density, ice floats on water. _______________________________________________________________________________ EXAMPLE 11.4 Intermolecular Forces in Liquids The viscosities of ether and water are 0.000233 and 0.00101 N s/m2 . Discuss their relative values in terms of the following molecular structures. O /\ water H ethyl ether H CH3 O C H 3 \/ \/ CH 2 CH2 •Method of Solution Molecules with strong intermolecular forces have greater viscosities than those that have weak intermolecular forces. Here both molecules are dipolar: water because of the O—H bonds, and ether because of the polar C—O bond. However, water has a higher viscosity than ether because of its ability to form hydrogen bonds between molecules. _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Intermolecular Forces and Liquids and Solids / 2 17 EXERCISES 6. What property of a liquid makes it possible to fill a glass of water even above the rim? 7. Which of the two liquids has the higher viscosity at room temperature? C 2 H5 OH or CH 3 OCH3 THE SOLID STATE STUDY OBJECTIVES 1. 2. 3. Determine the number of atoms per unit cell given the type of unit cell. Calculate the spacing between planes in a crystal using Bragg's equation. Predict whether the type of crystal formed by a substance is ionic, covalent, molecular, or metallic. Crystal Structure. The atoms, molecules, or ions of a c rystalline solid occupy specific positions in the solid, and possess long-range order. The u nit cell is the smallest unit that when repeated over and over again generates the entire crystal. Three types of unit cell are the simple cubic cell (scc), the body-centered cubic cell (bcc), and the face-centered cubic cell (fcc). These types are illustrated in Figure 11.17 of the text. In a s imple cubic cell , particles are located only at the corners of each unit cell. In a b ody-centered cubic cell , particles are located at the center of the cell as well as at the corners. In a f ace-centered cubic cell , particles are found at the center of each of the six faces of the cell as well as at the corners. In many calculations involving properties of a crystal, it is important to know how many atoms or ions are contained in each unit cell. Atoms at the corners of unit cells are shared by neighboring unit cells. For a cubic cell, each corner atom is shared by eight unit cells and is counted as 1/8 particle for each unit cell. A face-centered atom is shared by two unit cells and is counted as 1/2 of an atom for each unit cell. An atom located at the center belongs wholly to that unit cell. The packing efficiency is the percentage of the cell space occupied by spheres. The empty space is the interstices or "holes" between spheres. There are many substances whose atomic arrangement can be pictured as a result of packing together identical spheres so as to achieve maximum density. Many metallic elements and many molecular crystals display these "closest packed" structures. The text describes these two "closest packed" structures in Section 11.4. Alternating layers of spheres in ABABAB… fashion yields a hexagonal close-packed (hcp) structure, while alternating layers of spheres in the ABCABC… fashion generates the cubic close-packed (ccp) structure. In both, structures each sphere has a coordination number of 12 which is the maximum coordination number. For both structures the packing efficiency is 74%. X-Ray Diffraction of Crystals. Crystal structures are analyzed experimentally by the technique of X-ray diffraction. Diffraction is a wave property. When an X-ray beam encounters a single crystal, the beam is scattered from the crystal at only a few angles, rather than randomly. The observed angle θ depends on the spacing d between layers or planes of atoms in the crystal, the wavelength (λ ) of the X rays, and the reflection order (n = 1, 2, 3, …). These quantities are related by the Bragg equation: nλ = 2d sin The spacing (d) between atomic planes in crystals can be determined by using X rays of known λ and by experimental measurement of . Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 18 / Intermolecular Forces and Liquids and Solids Types of Crystals. In a solid, atoms, molecules, or ions occupy specific positions called lattice points. Crystalline solids are characterized by a regular three-dimensional arrangement of lattice points. In ionic solids the lattice points are occupied by positive and negative ions. In all ionic compounds there are continuous threedimensional networks of alternating positive and negative ions held together by strong electrostatic forces (ionic bonds). Most ionic crystals possess high lattice energies, high melting points, and high boiling points. In ionic solids there are no discrete molecules as are found in molecular or covalent substances. In covalent crystals the lattice points are occupied by atoms that are held by a network of covalent bonds. Diamond, graphite, and quartz are well known examples. Materials of this type have high melting points and are extremely hard because of the large number of covalent bonds that have to be broken to melt or break up the crystal. The entire crystal can be thought of as one giant molecule. Covalent compounds form crystals in which the lattice positions are occupied by molecules. Such solids are called molecular crystals. Molecular crystals are soft and have low melting points. These properties are the result of the relatively weak intermolecular forces (van der Waals forces and hydrogen bonding) that hold the molecules in the crystal. Crystals of polar compounds are held together by dipole-dipole forces and dispersion forces. Only dispersion forces occur in the lattices of nonpolar compounds. As a rule polar compounds melt at higher temperatures than nonpolar compounds of comparable molecular mass. Metallic crystals are quite strong. Most transition metals have high melting points and densities. In metals the array of lattice points is occupied by positive ions. The outer electrons of the metal atoms are loosely held and move freely from ion to ion throughout the metallic crystal. This mobility of electrons in metals accounts for one of the characteristic properties of metals, namely the ability to conduct electricity. Amorphous solids lack a well-defined arrangement and long range order. Glass is considered to be an amorphous solid because it lacks a regular three dimensional arrangement of atoms. _______________________________________________________________________________ EXAMPLE 11.5 Number of Atoms Per Unit Cell If atoms of a solid occupy a face-centered cubic lattice, how many atoms are there per unit cell. •Method of Solution In a face-centered cubic cell there are atoms at each of the eight corners, and there is one in each of the six faces. •Calculation 8 corners (1/8 atom per corner) + 6 faces (1/2 atom per face) = 4 atoms per unit cell _______________________________________________________________________________ EXAMPLE 11.6 Dimensions of a Unit Cell Potassium crystallizes in a body-centered cubic lattice, and has a density of 0.856 g/cm3 at 25°C. a. How many atoms are there per unit cell? b. What is the length of an edge of the cell? •Method of Solution a. A body-centered cubic structure has one K atom in the center, and eight other K atoms, one at each corner of the cube. The corner atoms, however, are shared by 8 adjoining cells. Each corner atom contributes 1/8 of an atom to the unit cell. The total number of K atoms per unit cell is 1+ Back Forward 1 (8) = 2 atoms 8 Main Menu TOC Study Guide TOC Textbook Website MHHE Website Intermolecular Forces and Liquids and Solids / 2 19 b. Ideally the crystal of potassium is made up of a large number of unit cells repeated over and over again. Thus, the density of the unit cell will be the same as the density of metallic K. Recall that density is an intensive property. density = mass volume mass of 2 K atoms a3 where a is the length of the side of the unit cell. density (unit cell) = •Calculation The mass of a K atom is: 39.1 g 1 m ol × = 6.50 × 10 –23 g/atom mol 6.02 × 1 0 23 a toms Substituting into the density equation yields: 2 ( 6.50 × 1 0 –23 g /atom) a3 a 3 = 1.52 × 10 –22 cm3 0.856 g/cm3 = Rearranging yields: a = 5.34 × 10 –8 cm = 534 pm _______________________________________________________________________________ EXAMPLE 11.7 Atomic Radius of a Potassium Atom Determine the radius of a K atom for the metallic potassium crystal in Example 11.8. •Method of Solution From Figure 11.22 in the text, it is clear that we need the diagonal distance of the body-centered cubic cell because only along the diagonal do the atoms actually touch. The diagonal c will equal 4r where r is the atomic radius. From the figure we see that c = 4r Therefore 4r = and c= 3a 3a where a is the length of an edge. Solving for r, we get r= a 3 4 = 534 pm × 3 4 r = 231 pm _______________________________________________________________________________ EXAMPLE 11.8 Using the Bragg Equation When X rays with a wavelength of 0.154 nm are diffracted by a crystal of metallic copper, the angle corresponding to the first order diffraction (n = 1) is found to be 37.06°. What is the spacing between planes of Cu atoms that gives rise to this diffraction angle? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 20 / Intermolecular Forces and Liquids and Solids •Method of Solution The spacing between Cu atoms corresponds to d in the Bragg equation, which we rearrange to solve for d. d= nλ 2 sin • Calculation For a first-order diffraction n = 1. The sine of 37.06° can be determined with your calculator. d= 0.154 nm 0 .154 nm = 2 sin 37.06° 2(0.603) = 0.128 nm (128 pm) _______________________________________________________________________________ EXAMPLE 11.9 Types of Crystals What type of force must be overcome in order to melt crystals of the following substances? a. Mg b. Cl 2 c . MgCl 2 d. SO2 e . Si •Method of Solution a. b. c. d. e. Magnesium is a metal and so it forms metallic crystals. Since the mobile electrons are shared between positive ions, the forces between Mg atoms could be described as covalent bonds, in this case called metallic bonds. The melting point is 1105°C. Cl 2 forms molecular crystals that are held together by intermolecular forces, more specifically dispersion forces. The melting point is –101°C. MgCl 2 is an ionic compound. When it melts, Mg2+ ions and Cl – ions break away from their lattice positions. To melt magnesium chloride, ionic bonds must be broken. The melting point is 1412°C. SO2 is a polar molecule. Thus, dipole-dipole attractions and dispersion forces must be overcome. The melting point is –73°C. Si crystallizes in a diamond structure with each silicon atom bound to four others by covalent bonds. It is a covalent crystal, and covalent bonds must be broken in order for it to melt. The melting point is 1410°C. _______________________________________________________________________________ EXERCISES 8. Nickel crystallizes in a face-centered cubic lattice with an edge length of 352 pm. Calculate the density of Ni. 9. Nickel crystallizes in a face-centered cubic lattice with an edge length of 352 pm. Given that the density of Ni is 8.94 g/cm3 , calculate Avogadro's number. 10. Copper crystallizes in a cubic system and the edge of the unit cell is 361 pm. The density of copper is 8.96 g/cm3 . How many atoms are contained in one unit cell? What type of cubic cell does Cu form? 11. Copper crystallizes in a face-centered cubic lattice. The density of copper is 8.96 g/cm3 . What is the length of the edge of a unit cell? 12. In a diffraction experiment, X rays of wavelength 0.154 nm were reflected from a gold crystal. The firstorder reflection was at 22.2°. What is the distance between the planes of gold atoms in pm? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Intermolecular Forces and Liquids and Solids / 2 21 13. The distance between the planes of gold atoms in a gold crystal is 204 pm. In a diffraction experiment the first-order reflection was observed at 22.2°. What was the wavelength of the X-rays? PHASE CHANGES AND PHASE DIAGRAMS STUDY OBJECTIVES 1. 2. 3. 4. 5. Describe the dynamic equilibrium between a liquid and its vapor state and list the factors that affect the vapor pressure of a liquid. Describe the significance of the critical point. Define the molar heats of fusion, vaporization, and sublimation. Using a phase diagram, predict what changes a substance will undergo as it is heated or cooled, or subjected to pressure. Use the Clausius-Clapeyron equation to relate vapor pressure, temperature, and the heat of vaporization of a liquid. Liquid-Vapor Equilibrium. Chemists refer to the different states of a substance that are present in a system as phases. A p hase is a homogeneous part of the system that is in contact with other parts of the system, but is separated by an noticeable boundary. Ice, liquid water, and water vapor can exist together in a suitable container. Each of these is a phase. When a liquid substance is placed in a closed container, it will not be long before molecules of this substance can be found in the gas phase above the liquid. Vaporization is the process in which liquids become gases. When a portion of a liquid evaporates the gaseous molecules exert a pressure called the vapor pressure. The term vapor is often applied to the gaseous state of a substance that is normally a liquid or solid at the temperature of interest. In a closed container the vapor pressure does not just continually increase, rather a state of equilibrium is reached in which the vapor pressure becomes constant and the amount of liquid remains constant. This equilibrium vapor pressure results from two opposing processes. The opposing process to vaporization is condensation. As the concentration of the molecules in the vapor phase increases, some will strike the liquid surface and condense. v aporization liquid vapor c ondensation When the two rates become equal, the vapor pressure remains constant and is called the e quilibrium vapor pressure , or just vapor pressure. Note that in the state of equilibrium, while the amounts of vapor and liquid do not change, there is considerable activity on the molecular level. Thus evaporation and condensation are constantly occurring, but at the same rates. Such an equilibrium state is referred to as a dynamic equilibrium. Vapor pressure is a function of the temperature. As temperature increases the evaporation rate increases, and so more molecules exist in the vapor phase at higher temperatures. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 22 / Intermolecular Forces and Liquids and Solids Table 11.4 Vapor Pressure of Water _________________________________ Temp (°C) Pressure (mm Hg) _________________________________ 20 17.54 30 31.82 40 55.32 50 92.51 80 355.1 86.5 460.0 100 760.0 _________________________________ Vapor Pressure (mm Hg) In a container open to the atmosphere, boiling of a liquid will occur when the temperature is raised high enough. In order for a bubble to form, the vapor inside the bubble must be able to push back the atmosphere. This will not occur until a temperature is reached at which the vapor pressure is greater than the atmospheric pressure. Therefore, the boiling point of a liquid is the temperature at which the vapor pressure is equal to the atmospheric pressure. Since the boiling point of a liquid depends on the atmospheric pressure, and the atmospheric pressure varies daily, the boiling point is not a constant. A normal boiling point is defined for purposes of comparing liquid substances. The normal boiling point is the boiling temperature when the external pressure is 1 atm. Figure 11.1 shows the vapor pressures and normal boiling points of three liquids. 152 0 CHCl 3 H2 O 760 0 0 50 61 10 0 Temperature (°C) Figure 11.1 The effect of temperature on the vapor pressure of chloroform and water. Notice that the normal boiling points for the liquids are the temperatures at which their vapor pressures are equal to 1 atm. The energy required to vaporize one mole of a liquid is called the molar heat of vaporization (∆ Hvap ). The value of ∆ Hvap is directly proportional to the strength of intermolecular forces. Liquids with relatively high heats of vaporization have low vapor pressures and high normal boiling points. The quantity ∆ Hvap can be determined experimentally. The vapor pressure of a liquid is proportional to the temperature. The quantitative relationship between the vapor pressure (P) and the absolute temperature (T) is given by the Clausius-Clapeyron equation ln P = – ∆ Hvap +C RT where ln is the natural logarithm, R is the ideal gas constant (8.314 J K–1 mol –1 ), and C is a constant. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Intermolecular Forces and Liquids and Solids / 2 23 A useful form of this equation is P1 ∆ Hvap T1 – T 2 ln = R P2 T1 T2 The ∆ Hvap can be calculated if the vapor pressures P1 and P2 are measured at two temperatures T1 and T2 . Alternatively if ∆ Hvap is already known and one vapor pressure P1 is known at temperature T 1 , you can calculate the vapor pressure P2 at some new temperature T2 . See Example 11.10. Boiling Point as a Measure of IMF. The boiling point and the heat of vaporization are convenient measures of the strength of intermolecular forces. In the liquid, molecules are very close together and are strongly influenced by intermolecular forces. In the gas phase, molecules are widely spaced, move rapidly, and have enough energy to overcome intermolecular forces. During vaporization, molecules are completely separated from one another. The heat of vaporization is the energy necessary to overcome intermolecular forces. The boiling point reflects the kinetic energy that liquid molecules must have to overcome intermolecular forces and escape into the gas phase. Critical Temperature and Pressure. When a liquid is heated in a closed container the vapor pressure increases, but boiling does not occur. The vapor cannot escape, and the vapor pressure continually rises. Eventually a temperature is reached at which the meniscus between liquid and vapor disappears. This temperature is called the critical temperature. The critical temperature is the highest temperature at which the substance can exist as a liquid. The critical pressure is the lowest pressure that will liquefy a gas at the critical temperature. Critical temperatures and pressures are listed in Table 11.7 (in the text). As we saw previously, intermolecular forces are dominant in liquids and effectively determine many of their properties. The critical temperature and pressure are no exceptions. Substances with high critical temperatures have strong intermolecular forces of attraction. Liquid-Solid Equilibrium. The temperature at which the solid and liquid are in dynamic equilibrium at 1 atm pressure is called the normal melting point or freezing point. Melting is also called fusion. During melting the average distance between molecules is increased slightly as evidenced by the approximately 10% decrease in density. fusion solid liquid freezing The energy required to melt one mole of a solid is called the molar heat of fusion, ∆ Hfus. Upon freezing, the substance will evolve the same amount of energy. The ∆ Hfus is always much less than ∆ Hvap because vaporization separates molecules completely from each other and so requires more energy than fusion. Heating Curves. A heating curve is a convenient way to summarize the solid-liquid-gas transitions for a compound. The heating curve of water is shown in Figure 11.2. It is the result of an experiment in which a given amount of ice, 1 mole for instance, at some initial temperature below 0°C is slowly heated at a constant rate. The curve shows that the temperature of ice increases on heating (line 1) until the melting point is reached. No rise in temperature occurs while ice is melting (line 2); as long as some ice remains, the temperature stays at 0°C. The length of line 2 is a measure of the heat necessary to melt 1 mole of ice, which is ∆ Hfus. Along line 3 the temperature of liquid water increases from 0°C to 100°C. The slope of the line ∆ t per joule of heat depends on the specific heat of liquid water. The greater slope for line 1 than for line 3 means that less heat is required to raise the temperature of ice than of water. This is evidenced by the difference in specific heats of ice and liquid water. The specific heat of ice is 2.09 J/g·°C versus 4.18 J/g·°C for liquid H2 O. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 24 / Intermolecular Forces and Liquids and Solids 4 Temperature (°C) 10 0 ∆ Hvap = 40.8 kJ/mol 5 3 2 ∆H fus = 6.0 kJ/mol 0 1 Heat absorbed (kJ) Figure 11.2 The heating curve for water. At 100°C the liquid begins to boil, and the heat added is used to bring about vaporization. The temperature does not rise until all the liquid has been transformed to gas. The length of line 4 is the heat required to vaporize 1 mole of liquid. Line 4 will always be longer than line 2 because ∆ Hvap > ∆ Hfus. Line 5 corresponds to the heating of steam. Again the slope of line 5 depends on the specific heat of steam, which is about 1.98 J/g·°C. Solid-Vapor Equilibrium. S ublimation is the evaporation of a solid directly into the vapor phase. Dry ice and iodine are substances that sublime readily. Ice also sublimes to some extent. As with liquids, the vapor pressure of a solid increases as the temperature increases. The direct conversion of solid to vapor is equivalent to melting the solid first and then vaporizing the liquid. From Hess's law we obtain: ∆ Hsub = ∆ Hfus + ∆ Hvap Pressure Phase Diagrams. From discussions in preceding sections we can see that the phase in which a substance exists depends on its temperature and pressure. In addition, two phases may exist in equilibrium at certain temperatures and pressures. Information about the stable phases for a specific compound is summarized by a phase diagram such as that shown in Figure 11.3. liquid solid gas Temperature Figure 11.3 A typical phase diagram.. A phase diagram is a graph of pressure versus temperature. The diagram is divided into three regions, one for the solid phase, one for the liquid phase, and one for the gas phase. The line between the solid and liquid Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Intermolecular Forces and Liquids and Solids / 2 25 regions is made up of points of P and T at which the solid and liquid phases are in equilibrium. The line between the solid and gas regions is made up of points of P and T at which solid and gas are in equilibrium. And the line between the liquid and gas regions gives temperatures and pressures at which liquid and gas are in equilibrium. Since all three lines intersect at the triple point, this one point describes the conditions under which gas, liquid, and solid are all in equilibrium. The phase diagram shows at a glance several properties of a substance: melting point, boiling point, and triple point. If a point (P and T) describing a system falls in the solid region, the substance exists as a solid. If the point falls on a line such as that between liquid and gas regions, the substance exists as liquid and vapor in equilibrium. ________________________________________________________________________________ EXAMPLE 11.10 Clausius-Clapeyron Equation The vapor pressure of water is 55.32 mm Hg at 40.0°C, and 92.51 mm Hg at 50.0°C. (See Table 5.3 in the text). Calculate the molar heat of vaporization of water. •Method of Solution The Clausius-Clapeyron equation relates the vapor pressure of a liquid to its molar heat of vaporization. ln P 1 = ∆ Hvap T1 – T 2 R P2 T1 T2 • Calculation Substituting the value given above ln ∆ Hvap 55.32 mm Hg = 313 K – 323 K 92.51 mm Hg 8.314 J K –1 m ol –1 (313 K)(323 K) ln 0.5980 = Error!) After taking the logarithm and rearranging yields: ∆ Hvap = – 0.5142(8.314 J K –1 m ol –1 )(1.01 × 1 0 5 K ) –10 K ∆ Hvap = 43,000 J/mol •Comment Compare this result to the ∆ Hvap listed in Table 11.7 of the text, 40,790 J/mol. The observed difference is real. The ∆ Hvap is less at the boiling point than it is at a lower temperature. Over the entire temperature range of a liquid, ∆ Hvap , is close to, but not really, a constant. _______________________________________________________________________________ EXAMPLE 11.11 Vapor Pressure and Boiling Point Using Figure 11.4 (in the text) estimate the boiling point of water in a pressure cooker at 2.0 atm pressure. •Method of Solution The boiling point of water at 2.0 atm pressure is the temperature at which the vapor pressure of water is 2.0 atm. According to the figure this should be approximately 120°C. _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 26 / Intermolecular Forces and Liquids and Solids _______________________________________________________________________________ EXAMPLE 11.12 Heat of Vaporization How much heat is evolved when 1.0 g of steam condenses at 100°C? •Method of Solution The heat of vaporization of water at 100°C is 40.8 kJ/mol. Condensation is the reverse of vaporization so it releases the same amount of heat. H2 O(g) → H2 O(l) ∆ H = –40.8 kJ/mol •Calculation The heat evolved when 1.0 g of water vapor condenses is: q = mol H 2 O × ∆ Hcond = 1.0 g × 1 m ol – 40.8 kJ × 18.0 g 1 mol = –2.27 kJ •Comment This is a lot of heat and shows us why burns from steam can be very serious. _______________________________________________________________________________ EXAMPLE 11.13 Heat of Fusion The heat of fusion of aluminum is 10.7 kJ/mol. How much energy is required to melt one ton of Al at the melting point, 660°C. •Method of Solution The heat of fusion is the energy required to melt 1 mole of a substance. Therefore we need to convert 2000 lb of Al to moles. •Calculation q = mol H2 O × ∆ Hfus 4 54 g 1 m ol Al 1 0.7 kJ × × = 360 × 10 3 kJ (360 MJ) 1 lb 27.0 g Al 1 mol Al _______________________________________________________________________________ EXAMPLE 11.14 The Critical Temperature = 2000 lb × Discuss the possibilities of liquefying oxygen and carbon dioxide at 25°C by increasing the pressure. •Method of Solution Above its critical temperature a substance cannot be liquefied by increasing the pressure. Substances having critical temperatures above 25°C can be liquefied at 25°C by application of sufficient pressure. With a critical temperature of 31°C, CO2 can be liquefied at 25°C with application of enough pressure. The pressure required can be read off the phase diagram for CO2 (Figure 11.41 in the text). Reading straight up from 25°C you will Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Intermolecular Forces and Liquids and Solids / 2 27 cross the liquid-vapor equilibrium point at a pressure of 67 atm. This is the pressure required to liquefy CO2 at 25°C. Oxygen has a critical temperature of –119°C; therefore, at a temperature above –119°C oxygen cannot be liquefied, no matter how much pressure is applied. To liquefy O2 , its temperature must be lowered below – 119°C and pressure applied. ___________________________________________________________________________ EXERCISES 14. The molar heat of vaporization of bromine (Br2 ) is 30.04 kJ/mol. What is the heat of vaporization per gram of bromine? 15. The heat of vaporization of iodine is 41.7 kJ/mol at the boiling point of iodine, 456 K. How much heat is required to vaporize 20.0 g of I2 ? 16. The molar heats of fusion and vaporization of potassium are 2.4 and 79.1 kJ/mol, respectively. Estimate the molar heat of sublimation of potassium metal. 17. The heat of vaporization of ammonia is 23.2 kJ/mol at its boiling point. How many grams of ammonia can be vaporized with 12.5 kJ of heat? 18. How much heat is needed to convert 1.0 kg of ice at 0°C completely into steam at 100°C? 19. Indicate whether each of the following processes is exothermic or endothermic. a. melting of ice b. evaporation of ethanol c. condensation of steam d. freezing of water e. sublimation of iodine 20. Which substance in each pair has the higher vapor pressure at a given temperature? a. C 2 H5 OH or CH 3 OH b. Cl 2 or Br2 c. CH 3 Br or CH3 Cl 21. The vapor pressure of liquid potassium is 1.00 mm Hg at 341°C and 10.0 mm Hg at 443°C. Calculate the molar heat of vaporization of potassium. 22. The vapor pressure of ethanol is 100 mm Hg at 34.9°C, and 400 mm Hg at 63.5°C. What is the molar heat of vaporization of ethanol? 23. The vapor pressure of liquid potassium is 10.0 mm Hg at 443°C. The heat of vaporization is 82.5 kJ/mol. Calculate the temperature where the vapor pressure is 760 mm Hg. This is the boiling point of liquid potassium. 24. The vapor pressure of ethanol is 400 mm Hg at 63.5°C. Its molar heat of vaporization is 41.7 kJ/mol. Calculate the vapor pressure of ethanol at 34.9°C. _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 28 / Intermolecular Forces and Liquids and Solids CONCEPTUAL QUESTIONS 1. Consider a steel cylinder containing 20 kg of CO2 in the form of liquid and gas. The valve is opened CO2 gas flows rapidly from the cylinder. Soon, however, the flow slows, and a coating of ice frost forms on the outside of the cylinder. When the valve is closed the cylinder is weighed and is found to still contain 14 kg of CO2 . a. Why is the flow faster at first? b. Why does the flow almost stop before the cylinder is empty? c. Why does the outside become icy? 2. Why do liquids boil at lower temperatures in the mountains as compared to sea level? 3. Explain how an increase in temperature causes an increase in vapor pressure. 4. Graphite can be transformed into diamond at high pressures. Which of these two forms of carbon has the higher density. PRACTICE TEST 1. 2. The enthalpies of vaporization of H2 S and H2 Se are 18.7 and 23.2 kJ/mol, respectively. Why is the ∆ Hvap of water (40.6 kJ/mol) so much higher than the values for H2 S and H2 Se? 3. Which member of each pair has the stronger intermolecular forces of attraction? a. CO 2 or SO 2 b . F 2 or Br2 c. H 2 O or H2 S d. HCl or C8 H18 4. In the laboratory, glassware is considered clean when water wets the glass in a continuous film. If droplets of water stand on the glass surface, the glass is considered dirty. Discuss the basis of this cleanliness test. 5. The viscosities of liquids generally decrease with increasing temperature. Water has the following viscosities in units of N s/m2 : 0.0018 at 0°C; 0.0010 at 20°C; 0.0005 at 55°C; and 0.0003 at 100°C. Interpret this trend on a molecular basis. 6. The meniscus for mercury in a glass tube is concave downward. Explain. 7. At what angle would a first-order reflection be observed in the diffraction of 0.090 nm X rays by a set of crystals planes for which d = 0.500 nm? 8. Calcium oxide like NaCl crystallizes in a face-centered cubic cell. a. How many Ca 2+ and O2– ions are in each unit cell? b. Given the ionic radius of Ca 2+ (99 pm) and O2– (140 pm), what is the length of an edge of the unit cell? c. Calculate the density of CaO. See Figure 11.22 in the textbook. 9. Back Identify the types of intermolecular forces for each of the following substances: a. H 2 O2 b. H 2 S c. SF6 d. NOCl e. NH3 From the following data calculate Avogadro's number. Potassium crystallizes in a body-centered cubic lattice with an edge of 530 pm. The density of K is 0.86 g/cm3 . Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Intermolecular Forces and Liquids and Solids / 2 29 10. Aluminum crystallizes in a face-centered cubic lattice with the length of an edge equal to 405 pm. Assume that a face-centered atom touches each of the surrounding corner atoms. Calculate the length of a diagonal of a face. 11. Pick the substance in each pair that has the higher melting point. a. ICl or KCl b. K or Fe c. Fe or Cl2 d. SiO 2 (quartz) or SiF4 e. CO 2 or SiO2 12. Distinguish between the boiling point of a liquid and the normal boiling point of a liquid. 13. Why does your skin feel cool when water evaporates from it? Why will you feel warmer on a more humid summer day than on a less humid day? 14. Use the following data to draw a rough phase diagram for methane: triple point –183.0°C; normal melting point –182.5°C; normal boiling point –161.5°C; critical point –83.0°C; critical pressure 45.6 atm. a. What is the stable state of methane at –160°C and 2.0 atm pressure? b. At a constant temperature of –182°C the pressure on a sample of methane is gradually increased from 1.0 mm Hg to 1.0 atm. What phases would be observed, and in what order would they occur? 15. Given the following data for N2 : Normal melting point: –210°C Specific heat of liquid: 2.0 J/g·°C Normal boiling point: –196°C Specific heat of gas: 1.0 J/g·°C Heat of fusion: 25 J/g Specific heat of solid: 1.6 J/g·°C Heat of vaporization: 200 J/g How much energy in kilojoules is required to convert 1000 g of N 2 at –206°C to N2 gas at 20°C? 16. In the following pairs indicate which member would have the higher heat of vaporization? a. He or Kr b. CH3 OH or CH 3 OCH3 17. Which one of the following compounds will exhibit hydrogen bonding? a. CH 4 b. HBr c. CH 3 OH d. CCl 4 18. The vapor pressure of mercury is 17.3 mm Hg at 200°C. What is its vapor pressure at 300°C if its molar heat of vaporization is 59.0 kJ/mol? ANSWERS Exercises 1. 2. 3. 4. 5. 6. 7. Back a. dipole-dipole forces and dispersion forces b. dispersion forces c. dispersion forces d. dipole-dipole forces and dispersion forces e. ionic bonds CS 2 because of its much larger number of electrons and therefore polarizability. b, c, and d. a. CO b. ICl c. H2 O d. AsH3 a. Hydrogen bonds and dispersion forces b. covalent bonds c. dispersion forces Surface tension C 2 H5 OH Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 30 / Intermolecular Forces and Liquids and Solids 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 8.94 g/cm3 6.02 × 10 23 4 atoms, face-centered cubic 361 pm 204 pm 0.154 nm 0.188 kJ/g 3.29 kJ 81.5 kJ/mol 9.16 g 3020 kJ a. endothermic b. endothermic a. CH 3 OH b. Cl 2 c. CH 3 Cl 82.5 kJ/mol ∆ Hvap = 41.7 kJ 1052 K 100 mm Hg c. exothermic d. exothermic e. endothermic Conceptual Questions 1. 2. 3. 4. a. The flow is faster at first because the vapor pressure of CO2 is higher at first. The temperature of the CO2 is higher at first. Therefore more of the CO2 is in the gas phase and the vapor pressure is higher at first. The liquid CO2 cools down because it supplies some of the heat of vaporization. b. The flow almost stops before the cylinder is empty because as the liqiud cools off the vapor pressure of CO2 drops so low that very little of the liquid CO2 can vaporize. c. As the liquid CO 2 evaporates the heat of vaporization must be continually supplied. Heat is taken form the steel cylinder and then from the surrounding air. As the temperature of the steel drops below the freezing point of water the water vapor in the air condenses (deposition) as ice on the cold cylinder. The atmospheric pressure decreases as altitude increases. Therefore the higher one goes the lower the vapor pressure needed to form the bubbles observed during boiling. The lower the vapor pressure needed for boiling, the lower the temperature. See Figure 11.32 in the text. The higher the temperature, the greater the kinetic energy of molecules and hence more molecules leave the surface by breaking intermolecular forces. Diamond Practice Test 1. 2. 3. 4. Back a. Dipole-dipole, dispersion forces, and hydrogen bonding b. Dipole-dipole and dispersion forces c. Dispersion forces d. Dipole-dipole and dispersion forces e. Dipole-dipole, dispersion forces and hydrogen bonding H2 O molecules are attracted to each other by hydrogen bonding, whereas H2 S and H2 Se are not. a. SO 2 b. Br2 c. H2 O d. C8 H18 Grease and water do not mix, thus if grease remains on the glass surface, water will bead up. When the surface is clean, the cohesive forces between water and glass cause water to spread out in a thin film and "wet" the glass. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Intermolecular Forces and Liquids and Solids / 2 31 5. Viscosity results from intermolecular forces. As the temperature of water is raised, more and more H bonds are broken and the viscosity decreases. 6. The cohesive forces between Hg atoms are greater than the adhesives forces between glass and mercury. 7. = 5.2° 8. a. 4 b. 478 pm c. 3.41 g/cm3 23 9. 6.10 × 10 10. 573 pm 11. a. KCl b. Fe c. Fe d. SiO2 e . SiO2 12. The bubble formation that signifies boiling cannot occur until the vapor pressure is great enough to push back the atmosphere. Thus the boiling point varies with atmospheric pressure. The normal boiling point of a liquid is the temperature at which its vapor pressure is 1.0 atm. 13. Energy is required for water to evaporate. When water evaporates from your skin, the heat of vaporization is absorbed from the skin, thus cooling it. You feel warmer on a humid day because the condensation of water vapor on your skin impedes the evaporation (cooling) process. 14. a. Liquid b. Gas, then liquid 15. 436 kJ 16. a. Kr b. CH3 OH 17. c. CH 3 OH 18. 237 mm Hg ___________________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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