Chapter #16 Liquids and Solids

Transcription

1 Chapter #16 Liquids and Solids 16.1 Intermolecular Forces 16.2 The Liquid State 16.3 An Introduction to Structures and Types of Solids 16.4 Structure and Bonding of Metals 16.5 Carbon and Silicon: Network Atomic Solids 16.6 Molecular Solids 16.7 Ionic Solids 16.8 Structures of Actual Ionic Solids 16.9 Lattice Defects Vapor Pressure and Changes of State Phase Diagrams Nanotechnology

2 Macroscopic Properties of Gases, Liquids, and Solids Condensed States of Matter Intermolecular Forces attractive forces between molecules

3 Macroscopic Properties of Gases, Liquids, and Solids

4 Phase Changes for Water and their Enthalpies D o sub Gas Condensation - D o vap= kj/mol -D o sub Sublimation Vaporization D o vap= 40.7 kj/mol Liquid Fusion/Freezing -D o fus = kj/mol Melting D o fus = 6.02 kj/mol Deposition Solid

5 vap = standard enthalpy of vaporization (or standard heat of vaporization) Example: 2 O (l) vaporization condensation 2 O (g) vap = 40.7 kj/mol Note: Standard pressure is 1 atm and the quantity of material is 1 mole

6 Vapor Pressure Recall: PV = nrt See Figures and At equilibrium, Evaporation rate = Condensation rate The pressure of the vapor at equilibrium is called the vapor pressure (P vap ) of the liquid (or solid) think of this as the number of gas molecules pushing down on the liquid or solid

7 Figure Measuring P vap if P vap is large substance called volatile" as intermolecular forces increase, P vap decreases

8 Vapor Pressures and Intermolecular Forces Figure More volatile substances: have higher P vap at lower T (more molecules in the vapor phase) Figure 16.48a TREND: higher P vap means weaker IMFs (molecules can more easily escape to the vapor phase)

9 Figure 16.47

10 Figure 16.48b From a P vap vs T plot (Fig 16.48a), how can we tell which substance has a smaller D vap? Let s plot the data another way Clausius-Clapeyron Equation: ln P D vap R 1 T C Similar to equation for the temperature dependence of the equilibrium constant (Zumdahl section 10.11)

11 Does the equation make sense? ln P D vap R 1 T C

12 Does the equation make sense? ln P D vap R 1 T C

13 Clausius-Clapeyron Equation: ln P D vap R 1 T C For 2 temperatures, T 1 and T 2, and vapor pressures, P 1 and P 2 : ln P P 1 D 1 T 1 2 vap 1 R T 2 From these equations, knowledge of any 2 pressures at 2 temperatures allows calculation of vap. Problem: Calculate the heat of vaporization of diethyl ether from the following vapor pressures: 400 mm g at 18 o C & 760 mm g at 35 o C. Plan: We are given P 1, P 2, T 1, and T 2 substitute into the C.C. equation and calculate the value of vap!

14 Changes of State:

15 Changes of State eating Curve q = C g DT = J/g q = C l DT = J/g q = D vap = kj/mol q = D fus = kj/mol q = C s DT = J/g Figure and Ch 9

16 Changes of State Melting point: Energy added disrupts the solid structure, increasing the potential energy of the molecules Solid and liquid states have the same P vap normal : total P vap = 1 atm Boiling point: P vap of a liquid becomes equal to the pressure of its environment normal : temperature at which P vap = 1 atm For water: 100 C

17 The Cooling Curve is Sensitive to Pressure Boiling occurs when the vapor pressure of a liquid becomes equal to the pressure of its environment. In the mountains, atmospheric pressure is reduced, so water boils at a lower temperature. (Less heat is required to achieve P vap = P atm ) Wouldn t it be great if we had a way to figure out in which physical state(s) a substance will exist at a given temperature and pressure?

18 (pure substances in closed containers) Phase diagrams are important "road maps" for understanding the effects of temperature and pressure on a substance

19 What determines the temperature at which liquids boil and solids melt? Intermolecular Forces Intramolecular forces: chemical bonds holding atoms together within a molecule ionic bonds, e.g. Na + Cl - ~ kj/mol covalent bonds, e.g. 2, C 4, etc. ~ kj/mol (the O- bond energy in 2 O is 500 kj/mol) metallic bonds, e.g. Au, Fe, etc. ~ kj/mol Inter molecular forces: forces between molecules Much weaker: ~ kj/mol

20 Intermolecular Forces (IMF) determine: Boiling Point/Melting Point Vapor Pressure Viscosity (measure of a liquid s resistance to flow) Surface Tension (resistance of a liquid to an increase in its surface area) Capillary Action (balance between cohesive and adhesive forces) Fig Fig Stronger IMFs require more energy to disrupt intermolecular interactions

21 Types of Intermolecular Forces Dipole-Dipole forces based on charge-charge interactions (electrostatic attratctions) occur in molecules with polar bonds and dipole moments typically 5-25 kj/mol Figure 16.2

22 ydrogen Bonding A special type of dipole-dipole force. A ydrogen atom is bound to the highly electronegative atoms O, N, or F. -X bonds are very polar A partially positive atom from one molecule is attracted to the partially negative X atom on another molecule (or to another highly electronegative atom, Z). This is the hydrogen bond. The atom sequence must be - Z : - X -, where both Z and X are one of the following three highly EN atoms: O, N, or F. The combined strength of many bonds can be large this is why water has the special properties that make it so important.

23 Water has particularly strong hydrogen bonding intermolecular forces. Isolated water molecule (gas phase) (+0.46) (-0.92) (+0.46) ydrogen bonding: Typically kj/mol Figure bonded water molecules (liquid or solid phase)

24 Boiling Points are igh in -bonded Liquids Exceptions Trend Figure 16.4 O, N, F small sizes and large electronegativities: strong -bonds

25 Identifying ydrogen Bonds Between Molecules - I Problem: In which of the following molecules does hydrogen bonding occur? Draw the hydrogen bonds where appropriate in the molecules. a) C 3 8 b) C 2 5 O c) Glycine: 2 NC 2 COO Plan: We examine each structure to see if F, N, or O is present, and if hydrogen can be bonded to them. Solution: a) for C 3 8, only C-C and C- bonds exist. No hydrogen bonds can be formed! C C C b) for C 2 5 O, the covalently bound to the oxygen can interact with the pair of electrons on the oxygen of another molecule to form a strong hydrogen bond. It can also hydrogen bond with water molecules in an aqueous solution of ethanol

26 C C O C C O C C N O O C C N O O C C N O O C C O O c) Glycine: b) Ethanol: Identifying ydrogen Bonds Between Molecules - II O

27 Adenine (A) Thymine (T) T A Guanine (G) Cytosine (C) C G

28 (London) Dispersion Forces Figure 16.5 Magnitude depends on polarizability and shape of the atom or molecule: heavier atoms/molecules with more electrons are generally more polarizable. Molecules with large surface areas interact more with their neighbors.

29 Rank by energy

30 There can be more than one type of intermolecular interaction between molecules/ions: Tetramethylammonium ion Nitrate ion ion-ion dispersion Ammonium ion Nitrate ion ion-ion hydrogen-bonding dispersion

31

32 Predicting the Types and Relative Strength of Intermolecular Forces Problem: Select the substance with the higher b.p. in each pair: a) C 3 Cl or C 3 O b) C 3 C 2 O or C 2 4 (O) 2 c) n-pentane (C 5 12 ) or neopentane (C 5 12 ) Plan: Examine the formulas and structures to determine the types of forces involved: Are ions present? Are the molecules polar or nonpolar? Is F, O, or N bound to? Do the molecules have different masses or shapes? Remember: a) Bonding (intramolecular) forces are stronger than intermolecular forces. b) ydrogen bonding is a strong type of dipole-dipole force. c) Dispersion forces are decisive when the major difference is molar mass or molecular shape.

33 Solutions: a) C 3 Cl (methyl chloride, MM=50.48g/mol) polar C 3 O (methanol, MM=32.04g/mol) polar and -bond ydrogen bonds are stronger than the dipole-dipole forces between the molecules in C 3 Cl. C 3 O will have the higher boiling point. Check: C 3 Cl (b.p. = C) and C 3 O (b.p. = C) b) C 3 C 2 O (ethyl alcohol, MM=46.07g/mol) and OC 2 C 2 O (ethylene glycol, MM=64.07g/mol) Both contain O- groups, but ethylene glycol is a di-hydroxy alcohol so it can have twice the hydrogen bonding. OC 2 C 2 O will have the higher boiling point. Check: C 3 C 2 O (b.p. = 78.5 C) and OC 2 C 2 O (b.p. = 290 C)

34 c) n-pentane (C 5 12 ) and neopentane (C 5 12 ) are both non-polar molecules with the same molecular mass. The tetrahedral neopentane makes less intermolecular contact than the straight-chain n-pentane. n-pentane should have greater dispersion forces and a slightly higher boiling point. Check: n-pentane (b.p.=36.1 C) and neopentane (b.p.=9.5 C)

35 Surface Tensions and Intermolecular Forces Surface tension: the resistance of a liquid to an increase in surface area; surface can be vapor/solid ; molecule at the surface only attracted by molecules next to and below it Figure 16.6 Large IMFs high surface tension

36 Adhesive Cohesive

37 Three examples to consider

38 Surfactants: Polar + Non-Polar ybrids Polar head Non-polar tail -C -CO - 2 -C 2 -C 2-2 as in soap -SO - 3 as in industrial detergents Air Air-liquid Air-water interface Weak non-covalent forces Polar liquid Surfactant: surface acting agent; lowers the surface tension of a liquid (easier spreading)

39 Three demos in this part of Ch 16 Liquid Nitrogen: there can be some blurry lines between liquids and solids; N 2 : m.p C, b.p -196 C At room temp flowers contain a lot of liquid water and rubber objects have properties more like slow-flowing liquids than solids. Only at much lower temperatures do these solid materials show the properties of solids. Sublimation of Dry Ice: the sublimation point of CO 2 is C at atmospheric pressure; a piece of dry ice is placed in a flask and a balloon is attached to the mouth of the flask to collect carbon dioxide gas. Phases of Bromine: molecular substances have relatively low melting and boiling points because the molecules are held together by fairly weak dispersion forces; Br 2 : m.p. -7 C, b.p 59 C can put it on dry ice that was below 195K (-78 C, -109 F) and let it warm up to room temp