General Chemistry 202 CHM202 General Information. General Chemistry 202 CHM202 Policies. General Chemistry 202 CHM202 Laboratory Guidelines

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1 General Chemistry 202 CHM202 General Information Instructor Meeting times and places Text and recommended materials Website Grading Schedule 1 General Chemistry 202 CHM202 Policies Equipment Instruction Method Study Procedures Evaluation Attendance Cheating Personal Technology 2 General Chemistry 202 CHM202 Laboratory Guidelines Attendance Safety Preparation General Procedures Data Recording Cleanup Reports 3 1

2 Frequently Asked Questions Q: Where does class meet on? A: It s in the syllabus. Q: May I leave class early tonight? A: I ll make a note of it. Q: I didn t do well on the test. When can I make it up? A: It s in the syllabus. 4 Frequently Asked Questions Q: I made a mistake on the test. May I have it back, please? A: No. Q: Is there a test on? A: It s in the syllabus. Q: What are your office hours? A: Posted on the class website. I will only be there by appointment. 5 Frequently Asked Questions Q: When can I make up the test I missed when I was absent? A: It s in the syllabus. Q: What if the absence wasn t excused? A: It s in the syllabus. Q: How do I get my absence excused? A: Offer me an excuse. (I may ask for documentation.) 6 2

3 Frequently Asked Questions Q: May I have extra time for my test? A: Please make arrangement through the proper channels. Such accomodations require documentation. Q: May I attend the earlier or later lab? A: No. Q: May I attend the earlier lecture? A: By request, but space is limited 7 Intermolecular Forces and Liquids and Solids Chapter 11 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 8 A phase is a homogeneous part of the system in contact with other parts of the system but separated from them by a welldefined boundary. 2 Phases Solid phase - ice Liquid phase - water 9 3

4 Intermolecular Forces Intermolecular forces are attractive forces between molecules. Intramolecular forces hold atoms together in a molecule. Intermolecular vs Intramolecular 41 kj to vaporize 1 mole of water (inter) 930 kj to break all O-H bonds in 1 mole of water (intra) Generally, intermolecular forces are much weaker than intramolecular forces. Measure of intermolecular force boiling point melting point DH vap DH fus DH sub 10 Intermolecular Forces Dipole-Dipole Forces Attractive forces between polar molecules Orientation of Polar Molecules in a Solid 11 Intermolecular Forces Ion-Dipole Forces Attractive forces between an ion and a polar molecule Ion-Dipole Interaction 12 4

5 Interaction Between Water and Cations in solution 13 Dispersion Forces Intermolecular Forces Attractive forces that arise as a result of temporary dipoles induced in atoms or molecules ion-induced dipole interaction dipole-induced dipole interaction 14 Induced Dipoles Interacting With Each Other 15 5

6 Dispersion Forces Continued Intermolecular Forces Polarizability is the ease with which the electron distribution in the atom or molecule can be distorted. Polarizability increases with: greater number of electrons more diffuse electron cloud Dispersion forces usually increase with molar mass. 16 Example 11.1 What type(s) of intermolecular forces exist between the following pairs? (a) HBr and H 2 S (b) Cl 2 and CBr 4 (c) I 2 and (d) NH 3 and C 6 H 6 Example 11.1 Strategy Classify the species into three categories: ionic, polar (possessing a dipole moment), and nonpolar. Keep in mind that dispersion forces exist between all species. Solutions (a) Both HBr and H 2 S are polar molecules. Therefore, the intermolecular forces present are dipole-dipole forces, as well as dispersion forces. (b) Both Cl 2 and CBr 4 are nonpolar, so there are only dispersion forces between these molecules. 6

7 Example 11.1 (c) I 2 is a homonuclear diatomic molecule and therefore nonpolar, so the forces between it and the ion,, are ion-induced dipole forces and dispersion forces. (d) NH 3 is polar, and C 6 H 6 is nonpolar. The forces are dipoleinduced dipole forces and dispersion forces. Hydrogen Bond Intermolecular Forces The hydrogen bond is a special dipole-dipole interaction between the hydrogen atom in a polar N-H, O-H, or F-H bond and an electronegative O, N, or F atom. A H B or A H A A & B are N, O, or F 20 Hydrogen Bond HCOOH and water 21 7

8 Why is the hydrogen bond considered a special dipole-dipole interaction? Decreasing molar mass Decreasing boiling point 22 Example 11.2 Which of the following can form hydrogen bonds with water? a) CH 3 OCH 3 b) CH 4 c) F 2 d) HCOOH e) Na + Example 11.2 Strategy A species can form hydrogen bonds with water if it contains one of the three electronegative elements (F, O, or N) or it has a H atom bonded to one of these three elements. Solution There are no electronegative elements (F, O, or N) in either CH 4 or Na +. Therefore, only CH 3 OCH 3, F 2, and HCOOH can form hydrogen bonds with water. 8

9 Example 11.2 Check Note that HCOOH (formic acid) can form hydrogen bonds with water in two different ways. HCOOH forms hydrogen bonds with two H 2 O molecules. Properties of Liquids Surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area. Strong intermolecular forces High surface tension 26 Properties of Liquids Cohesion is the intermolecular attraction between like molecules Adhesion is an attraction between unlike molecules Adhesion Cohesion 27 9

10 Properties of Liquids Viscosity is a measure of a fluid s resistance to flow. Strong intermolecular forces High viscosity 28 3-D Structure of Water Water is a Unique Substance Maximum Density 4 0 C Density of Water Ice is less dense than water 29 A crystalline solid possesses rigid and long-range order. In a crystalline solid, atoms, molecules or ions occupy specific (predictable) positions. An amorphous solid does not possess a well-defined arrangement and long-range molecular order. A unit cell is the basic repeating structural unit of a crystalline solid. Unit Cell lattice point Unit cells in 3 dimensions At lattice points: Atoms Molecules Ions 30 10

11 Seven Basic Unit Cells 31 Three Types of Cubic Unit Cells 32 Arrangement of Identical Spheres in a Simple Cubic Cell 33 11

12 Arrangement of Identical Spheres in a Body-Centered Cubic Cell 34 A Corner Atom, an Edge-Centered Atom and a Face-Centered Atom Shared by 8 unit cells Shared by 4 unit cells Shared by 2 unit cells 35 Number of Atoms Per Unit Cell 1 atom/unit cell (8 x 1/8 = 1) 2 atoms/unit cell (8 x 1/8 + 1 = 2) 4 atoms/unit cell (8 x 1/8 + 6 x 1/2 = 4) 36 12

13 Relation Between Edge Length and Atomic Radius 37 Closest Packing: Hexagonal and Cubic hexagonal cubic 38 Exploded Views 39 13

14 Example 11.3 Gold (Au) crystallizes in a cubic close-packed structure (the face-centered cubic unit cell) and has a density of 19.3 g/cm 3. Calculate the atomic radius of gold in picometers. Example 11.3 Strategy We want to calculate the radius of a gold atom. For a face-centered cubic unit cell, the relationship between radius (r) and edge length (a), according to Figure 11.22, is. Therefore, to determine r of a Au atom, we need to find a. The volume of a cube is. Thus, if we can determine the volume of the unit cell, we can calculate a. We are given the density in the problem. Example 11.3 The sequence of steps is summarized as follows: Solution Step 1: We know the density, so in order to determine the volume, we find the mass of the unit cell. Each unit cell has eight corners and six faces. The total number of atoms within such a cell, according to Figure 11.19, is 14

15 Example 11.3 The mass of a unit cell in grams is From the definition of density (d = m/v), we calculate the volume of the unit cell as follows: Example 11.3 Step 2: Because volume is length cubed, we take the cubic root of the volume of the unit cell to obtain the edge length (a) of the cell Step 3: From Figure we see that the radius of an Au sphere (r) is related to the edge length by Example 11.3 Therefore, = 144 pm 15

16 An Arrangement for Obtaining the X-ray Diffraction Pattern of a Crystal. 46 Reflection of X-rays from Two Layers of Atoms Extra distance = BC + CD = 2d sinq = nl (Bragg Equation) 47 Example 11.4 X rays of wavelength nm strike an aluminum crystal; the rays are reflected at an angle of Assuming that n = 1, calculate the spacing between the planes of aluminum atoms (in pm) that is responsible for this angle of reflection. The conversion factor is obtained from 1 nm = 1000 pm. 16

17 Example 11.4 Strategy This is an application of Equation (11.1). Solution Converting the wavelength to picometers and using the angle of reflection (19.3 ), we write Types of Crystals Ionic Crystals Lattice points occupied by cations and anions Held together by electrostatic attraction Hard, brittle, high melting point Poor conductor of heat and electricity CsCl ZnS CaF 2 50 Example 11.5 How many Na + and Cl ions are in each NaCl unit cell? 17

18 Example 11.5 Solution NaCl has a structure based on a face-centered cubic lattice. One whole Na + ion is at the center of the unit cell, and there are twelve Na + ions at the edges. Because each edge Na + ion is shared by four unit cells, the total number of Na + ions is 1 + (12 ¼) = 4. Similarly, there are six Cl ions at the face centers and eight Cl ions at the corners. Each face-centered ion is shared by two unit cells, and each corner ion is shared by eight unit cells, so the total number of Cl ions is (6 ½) + (8 1/8) = 4. Thus, there are four Na + ions and four Cl ions in each NaCl unit cell. Check This result agrees with sodium chloride s empirical formula. Example 11.6 The edge length of the NaCl unit cell is 564 pm. What is the density of NaCl in g/cm 3? Example 11.6 Strategy To calculate the density, we need to know the mass of the unit cell. The volume can be calculated from the given edge length because V = a 3. How many Na + and Cl ions are in a unit cell? What is the total mass in amu? What are the conversion factors between amu and g and between pm and cm? Solution From Example 11.5 we see that there are four Na + ions and four Cl ions in each unit cell. So the total mass (in amu) of a unit cell is mass = 4(22.99 amu amu) = amu 18

19 Example 11.6 Converting amu to grams, we write The volume of the unit cell is V = a 3 = (564 pm) 3. Converting pm 3 to cm 3, the volume is given by Finally, from the definition of density = 2.16 g/cm 3 Types of Crystals Covalent Crystals Lattice points occupied by atoms Held together by covalent bonds Hard, high melting point Poor conductor of heat and electricity carbon atoms diamond graphite 56 Types of Crystals Molecular Crystals Lattice points occupied by molecules Held together by intermolecular forces Soft, low melting point Poor conductor of heat and electricity water benzene 57 19

20 Types of Crystals Metallic Crystals Lattice points occupied by metal atoms Held together by metallic bonds Soft to hard, low to high melting point Good conductors of heat and electricity nucleus & inner shell e - mobile sea of e - Cross Section of a Metallic Crystal 58 Crystal Structures of Metals 59 Types of Crystals 60 20

21 An amorphous solid does not possess a well-defined arrangement and long-range molecular order. A glass is an optically transparent fusion product of inorganic materials that has cooled to a rigid state without crystallizing Crystalline quartz (SiO 2 ) Non-crystalline quartz glass 61 Phase Changes Least Order Greatest Order 62 Effect of Temperature on Kinetic Energy T 2 > T

22 The equilibrium vapor pressure is the vapor pressure measured when a dynamic equilibrium exists between condensation and evaporation H 2 O (l) H 2 O (g) Dynamic Equilibrium Rate of condensation = Rate of evaporation 64 Measurement of Vapor Pressure Before Evaporation At Equilibrium 65 Molar heat of vaporization (DH vap ) is the energy required to vaporize 1 mole of a liquid at its boiling point. Clausius-Clapeyron Equation ln P = - DH vap RT + C P = (equilibrium) vapor pressure T = temperature (K) R = gas constant (8.314 J/K mol) Vapor Pressure Versus Temperature 66 22

23 Alternate Forms of the Clausius-Clapeyron Equation At two temperatures or 67 Example 11.7 Diethyl ether is a volatile, highly flammable organic liquid that is used mainly as a solvent. The vapor pressure of diethyl ether is 401 mmhg at 18 C. Calculate its vapor pressure at 32 C. Example 11.7 Strategy We are given the vapor pressure of diethyl ether at one temperature and asked to find the pressure at another temperature. Therefore, we need Equation (11.5). Solution Table 11.6 tells us that DH vap = 26.0 kj/mol. The data are From Equation (11.5) we have 23

24 Example 11.7 Taking the antilog of both sides (see Appendix 4), we obtain Hence P 2 = 656 mmhg Check We expect the vapor pressure to be greater at the higher temperature. Therefore, the answer is reasonable. The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure. The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm. 71 The critical temperature (T c ) is the temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure. The critical pressure (P c ) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature

25 The Critical Phenomenon of SF 6 T < T c T > T c T ~ T c T < T c 73 Solid-Liquid Equilibrium H 2 O (s) H 2 O (l) The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium. 74 Molar heat of fusion (DH fus ) is the energy required to melt 1 mole of a solid substance at its freezing point

26 Heating Curve 76 Solid-Gas Equilibrium H 2 O (s) H 2 O (g) Molar heat of sublimation (DH sub ) is the energy required to sublime 1 mole of a solid. DH sub = DH fus + DH vap ( Hess s Law) 77 Example 11.8 Calculate the amount of energy (in kilojoules) needed to heat 346 g of liquid water from 0 C to 182 C. Assume that the specific heat of water is J/g C over the entire liquid range and that the specific heat of steam is 1.99 J/g C. 26

27 Example 11.8 Strategy The heat change (q) at each stage is given by q = msdt (see p. 247), where m is the mass of water, s is the specific heat, and Dt is the temperature change. If there is a phase change, such as vaporization, then q is given by ndh vap, where n is the number of moles of water. Solution The calculation can be broken down in three steps. Step 1: Heating water from 0 C to 100 C Using Equation (6.12) we write Example 11.8 Step 2: Evaporating 346 g of water at 100 C (a phase change) In Table 11.6 we see DH vap = kj/mol for water, so Step 3: Heating steam from 100 C to 182 C Example 11.8 The overall energy required is given by Check All the qs have a positive sign, which is consistent with the fact that heat is absorbed to raise the temperature from 0 C to 182 C. Also, as expected, much more heat is absorbed during the phase transition. 27

28 A phase diagram summarizes the conditions at which a substance exists as a solid, liquid, or gas. Phase Diagram of Water 82 Phase Diagram of Carbon Dioxide At 1 atm CO 2 (s) CO 2 (g) 83 Effect of Increase in Pressure on the Melting Point of Ice and the Boiling Point of Water 84 28