1. © 2012 Pearson Education, Inc. Chapter 11 Liquids and Intermolecular Forces John D. Bookstaver St. Charles Community College Cottleville, MO Lecture Presentation

2. Intermolecular Forces © 2012 Pearson Education, Inc. States of Matter The fundamental difference between states of matter is the distance between particles.

3. Intermolecular Forces © 2012 Pearson Education, Inc. States of Matter Because in the solid and liquid states particles are closer together, we refer to them as condensed phases.

4. Intermolecular Forces © 2012 Pearson Education, Inc. The States of Matter The state a substance is in at a particular temperature and pressure depends on two antagonistic entities: –The kinetic energy of the particles. –The strength of the attractions between the particles.

5. Intermolecular Forces © 2012 Pearson Education, Inc. Intermolecular Forces The attractions between molecules are not nearly as strong as the intramolecular attractions that hold compounds together.

6. Intermolecular Forces © 2012 Pearson Education, Inc. Intermolecular Forces These intermolecular attractions are, however, strong enough to control physical properties, such as boiling and melting points, vapor pressures, and viscosities.

7. Intermolecular Forces © 2012 Pearson Education, Inc. Intermolecular Forces These intermolecular forces as a group are referred to as van der Waals forces.

8. Intermolecular Forces © 2012 Pearson Education, Inc. van der Waals Forces Dipole–dipole interactions Hydrogen bonding London dispersion forces

9. Intermolecular Forces © 2012 Pearson Education, Inc. London Dispersion Forces While the electrons in the 1s orbital of helium would repel each other (and, therefore, tend to stay far away from each other), it does happen that they occasionally wind up on the same side of the atom.

10. Intermolecular Forces © 2012 Pearson Education, Inc. London Dispersion Forces At that instant, then, the helium atom is polar, with an excess of electrons on the left side and a shortage on the right side.

11. Intermolecular Forces © 2012 Pearson Education, Inc. London Dispersion Forces Another helium atom nearby, then, would have a dipole induced in it, as the electrons on the left side of helium atom 2 repel the electrons in the cloud on helium atom 1.

12. Intermolecular Forces © 2012 Pearson Education, Inc. London Dispersion Forces London dispersion forces, or dispersion forces, are attractions between an instantaneous dipole and an induced dipole.

13. Intermolecular Forces © 2012 Pearson Education, Inc. London Dispersion Forces These forces are present in all molecules, whether they are polar or nonpolar. The tendency of an electron cloud to distort in this way is called polarizability.

14. Intermolecular Forces © 2012 Pearson Education, Inc. Factors Affecting London Forces The shape of the molecule affects the strength of dispersion forces: long, skinny molecules (like n-pentane) tend to have stronger dispersion forces than short, fat ones (like neopentane). This is due to the increased surface area in n-pentane.

15. Intermolecular Forces © 2012 Pearson Education, Inc. Factors Affecting London Forces The strength of dispersion forces tends to increase with increased molecular weight. Larger atoms have larger electron clouds that are easier to polarize.

16. Intermolecular Forces © 2012 Pearson Education, Inc. Dipole–Dipole Interactions Molecules that have permanent dipoles are attracted to each other. –The positive end of one is attracted to the negative end of the other, and vice versa. –These forces are only important when the molecules are close to each other.

17. Intermolecular Forces © 2012 Pearson Education, Inc. Dipole–Dipole Interactions The more polar the molecule, the higher its boiling point.

18. Intermolecular Forces © 2012 Pearson Education, Inc. Which Have a Greater Effect? Dipole–Dipole Interactions or Dispersion Forces If two molecules are of comparable size and shape, dipole–dipole interactions will likely be the dominating force. If one molecule is much larger than another, dispersion forces will likely determine its physical properties.

19. Intermolecular Forces © 2012 Pearson Education, Inc. How Do We Explain This? The nonpolar series (SnH 4 to CH 4 ) follow the expected trend. The polar series follow the trend until you get to the smallest molecules in each group.

20. Intermolecular Forces © 2012 Pearson Education, Inc. Hydrogen Bonding The dipole–dipole interactions experienced when H is bonded to N, O, or F are unusually strong. We call these interactions hydrogen bonds.

21. Intermolecular Forces © 2012 Pearson Education, Inc. Hydrogen Bonding Hydrogen bonding arises in part from the high electronegativity of nitrogen, oxygen, and fluorine. Also, when hydrogen is bonded to one of those very electronegative elements, the hydrogen nucleus is exposed.

22. Intermolecular Forces © 2012 Pearson Education, Inc. Sample Exercise 11.1 Identifying Substances That Can Form Hydrogen Bonds Solution Analyze We are given the chemical formulas of four compounds and asked to predict whether they can participate in hydrogen bonding. All the compounds contain H, but hydrogen bonding usually occurs only when the hydrogen is covalently bonded to N, O, or F. Plan We analyze each formula to see if it contains N, O, or F directly bonded to H. There also needs to be a nonbonding pair of electrons on an electronegative atom (usually N, O, or F) in a nearby molecule, which can be revealed by drawing the Lewis structure for the molecule. Solve The foregoing criteria eliminate CH 4 and H 2 S, which do not contain H bonded to N, O, or F. They also eliminate CH 3 F, whose Lewis structure shows a central C atom surrounded by three H atoms and an F atom. (Carbon always forms four bonds, whereas hydrogen and fluorine form one each.) Because the molecule contains a C  F bond and not a H  F bond, it does not form hydrogen bonds. In H 2 NNH 2, however, we find N  H bonds, and the Lewis structure shows a nonbonding pair of electrons on each N atom, telling us hydrogen bonds can exist between the molecules: In which of these substances is hydrogen bonding likely to play an important role in determining physical properties: methane (CH 4 ), hydrazine (H 2 NNH 2 ), methyl fluoride (CH 3 F), hydrogen sulfide (H 2 S)?

23. Intermolecular Forces © 2012 Pearson Education, Inc. Continued Check Although we can generally identify substances that participate in hydrogen bonding based on their containing N, O, or F covalently bonded to H, drawing the Lewis structure for the interaction provides a way to check the prediction. Practice Exercise In which of these substances is significant hydrogen bonding possible: methylene chloride (CH 2 Cl 2 ), phosphine (PH 3 ), hydrogen peroxide (HOOH), acetone (CH 3 COCH 3 )? Answer: HOOH Sample Exercise 11.1 Identifying Substances That Can Form Hydrogen Bonds

24. Intermolecular Forces © 2012 Pearson Education, Inc. Ion–Dipole Interactions Ion–dipole interactions (a fourth type of force) are important in solutions of ions. The strength of these forces is what makes it possible for ionic substances to dissolve in polar solvents.

25. Intermolecular Forces © 2012 Pearson Education, Inc. Summarizing Intermolecular Forces

26. Intermolecular Forces © 2012 Pearson Education, Inc. Sample Exercise 11.2 Predicting Types and Relative Strengths of Intermolecular Attractions Solution Analyze We need to assess the intermolecular forces in these substances and use that information to determine the relative boiling points. Plan The boiling point depends in part on the attractive forces in each substance. We need to order these according to the relative strengths of the different kinds of intermolecular attractions. Solve The attractive forces are stronger for ionic substances than for molecular ones, so BaCl 2 should have the highest boiling point. The intermolecular forces of the remaining substances depend on molecular weight, polarity, and hydrogen bonding. The molecular weights are H 2 (2), CO (28), HF (20), and Ne (20). The boiling point of H 2 should be the lowest because it is nonpolar and has the lowest molecular weight. The molecular weights of CO, HF, and Ne are similar. Because HF can hydrogen bond, however, it should have the highest boiling point of the three. Next is CO, which is slightly polar and has the highest molecular weight. Finally, Ne, which is nonpolar, should have the lowest boiling point of these three. The predicted order of boiling points is, therefore, H 2 < Ne < CO < HF < BaCl 2 Check The boiling points reported in the literature are H 2 (20 K), Ne (27 K), CO (83 K), HF (293 K), and BaCl 2 (1813 K)—in agreement with our predictions. List the substances BaCl 2, H 2, CO, HF, and Ne in order of increasing boiling point.

27. Intermolecular Forces © 2012 Pearson Education, Inc. Sample Exercise 11.2 Predicting Types and Relative Strengths of Intermolecular Attractions Continued (a) Identify the intermolecular attractions present in the following substances, and (b) select the substance with the highest boiling point: CH 3 CH 3, CH 3 OH, and CH 3 CH 2 OH. Answers: (a) CH 3 CH 3 has only dispersion forces, whereas the other two substances have both dispersion forces and hydrogen bonds, (b) CH 3 CH 2 OH Practice Exercise

28. Intermolecular Forces © 2012 Pearson Education, Inc. Intermolecular Forces Affect Many Physical Properties The strength of the attractions between particles can greatly affect the properties of a substance or solution.

29. Intermolecular Forces © 2012 Pearson Education, Inc. Viscosity Resistance of a liquid to flow is called viscosity. It is related to the ease with which molecules can move past each other. Viscosity increases with stronger intermolecular forces and decreases with higher temperature.

30. Intermolecular Forces © 2012 Pearson Education, Inc. Surface Tension Surface tension results from the net inward force experienced by the molecules on the surface of a liquid.

31. Intermolecular Forces © 2012 Pearson Education, Inc. Phase Changes

32. Intermolecular Forces © 2012 Pearson Education, Inc. Energy Changes Associated with Changes of State The heat of fusion is the energy required to change a solid at its melting point to a liquid.

33. Intermolecular Forces © 2012 Pearson Education, Inc. Energy Changes Associated with Changes of State The heat of vaporization is defined as the energy required to change a liquid at its boiling point to a gas.

34. Intermolecular Forces © 2012 Pearson Education, Inc. Energy Changes Associated with Changes of State The heat of sublimation is defined as the energy required to change a solid directly to a gas.

35. Intermolecular Forces © 2012 Pearson Education, Inc. Energy Changes Associated with Changes of State The heat added to the system at the melting and boiling points goes into pulling the molecules farther apart from each other. The temperature of the substance does not rise during a phase change.

36. Intermolecular Forces © 2012 Pearson Education, Inc. Sample Exercise 11.3 Calculating  H for Temperature and Phase Changes Solution Analyze Our goal is to calculate the total heat required to convert 1 mol of ice at  25  C to steam at 125  C. Plan We can calculate the enthalpy change for each segment and then sum them to get the total enthalpy change (Hess’s law, Section 5.6). Calculate the enthalpy change upon converting 1.00 mol of ice at  25  C to steam at 125  C under a constant pressure of 1 atm. The specific heats of ice, liquid water, and steam are 2.03 J/g-K, 4.18 J/g-K, and 1.84 J/g-K, respectively. For H 2 O,  H fus = 6.01 kJ/mol and  H vap = 40.67 kJ/mol. AB:  H = (1.00 mol)(18.0 g/mol)(2.03 J/g-K)(25 K) = 914 J = 0.91 kJ Solve For segment AB in Figure 11.22, we are adding enough heat to ice to increase its temperature by 25  C. A temperature change of 25  C is the same as a temperature change of 25 K, so we can use the specific heat of ice to calculate the enthalpy change during this process:

37. Intermolecular Forces © 2012 Pearson Education, Inc. Continued For segment BC in Figure 11.22, in which we convert ice to water at 0  C, we can use the molar enthalpy of fusion directly: The enthalpy changes for segments CD, DE, and EF can be calculated in similar fashion: The total enthalpy change is the sum of the changes of the individual steps: CD:  H = (1.00 mol)(18.0 g/mol)(4.18 J/g-K)(100 K) = 7520 J = 7.52 kJ DE:  H = (1.00 mol)(40.67 kJ/mol) = 40.7 kJ EF:  H = (1.00 mol)(18.0 g/mol)(1.84 J/g-K)(25 K) = 830 J = 0.83 kJ  H = 0.91 kJ + 6.01 kJ + 7.52 kJ + 40.7 kJ + 0.83 kJ = 56.0 kJ BC:  H = (1.00 mol)(6.01 kJ/mol) = 6.01 kJ Sample Exercise 11.3 Calculating  H for Temperature and Phase Changes

38. Intermolecular Forces © 2012 Pearson Education, Inc. Continued Check The components of the total energy change are reasonable relative to the horizontal lengths (heat added) of the segments in Figure 11.22. Notice that the largest component is the heat of vaporization. Practice Exercise What is the enthalpy change during the process in which 100.0 g of water at 50.0  C is cooled to ice at  30.0  C? (Use the specific heats and enthalpies for phase changes given in Sample Exercise 11.3.) Answer:  20.9 kJ  33.4 kJ  6.09 kJ =  60.4 kJ Sample Exercise 11.3 Calculating  H for Temperature and Phase Changes

39. Intermolecular Forces © 2012 Pearson Education, Inc. Vapor Pressure At any temperature some molecules in a liquid have enough energy to break free. As the temperature rises, the fraction of molecules that have enough energy to break free increases.

40. Intermolecular Forces © 2012 Pearson Education, Inc. Vapor Pressure As more molecules escape the liquid, the pressure they exert increases.

41. Intermolecular Forces © 2012 Pearson Education, Inc. Vapor Pressure The liquid and vapor reach a state of dynamic equilibrium: liquid molecules evaporate and vapor molecules condense at the same rate.

42. Intermolecular Forces © 2012 Pearson Education, Inc. Vapor Pressure The boiling point of a liquid is the temperature at which its vapor pressure equals atmospheric pressure. The normal boiling point is the temperature at which its vapor pressure is 760 torr.

43. Intermolecular Forces © 2012 Pearson Education, Inc. Vapor Pressure The natural log of the vapor pressure of a liquid is inversely proportional to its temperature.

44. Intermolecular Forces © 2012 Pearson Education, Inc. Vapor Pressure This relationship is quantified in the Clausius–Clapeyron equation: ln P = −  H vap /RT + C, where C is a constant.

45. Intermolecular Forces © 2012 Pearson Education, Inc. Sample Exercise 11.4 Relating Boiling Point to Vapor Pressure Solution Analyze We are asked to read a graph of vapor pressure versus temperature to determine the boiling point of a substance at a particular pressure. The boiling point is the temperature at which the vapor pressure is equal to the external pressure. Plan We need to convert 0.80 atm to torr because that is the pressure scale on the graph. We estimate the location of that pressure on the graph, move horizontally to the vapor pressure curve, and then drop vertically from the curve to estimate the temperature. Use Figure 11.25 to estimate the boiling point of diethyl ether under an external pressure of 0.80 atm.

46. Intermolecular Forces © 2012 Pearson Education, Inc. Sample Exercise 11.4 Relating Boiling Point to Vapor Pressure Solve The pressure equals (0.80 atm)(760 torr⁄atm) = 610 torr. From Figure 11.25 we see that the boiling point at this pressure is about 27  C, which is close to room temperature. Comment We can make a flask of diethyl ether boil at room temperature by using a vacuum pump to lower the pressure above the liquid to about 0.8 atm. Practice Exercise At what external pressure will ethanol have a boiling point of 60  C? Answer: about 340 torr (0.45 atm) Continued

47. Intermolecular Forces © 2012 Pearson Education, Inc. Phase Diagrams Phase diagrams display the state of a substance at various pressures and temperatures, and the places where equilibria exist between phases.

48. Intermolecular Forces © 2012 Pearson Education, Inc. Phase Diagrams The liquid–vapor interface starts at the triple point (T), at which all three states are in equilibrium, and ends at the critical point (C), above which the liquid and vapor are indistinguishable from each other.

49. Intermolecular Forces © 2012 Pearson Education, Inc. Phase Diagrams Each point along this line is the boiling point of the substance at that pressure.

50. Intermolecular Forces © 2012 Pearson Education, Inc. Phase Diagrams The interface between liquid and solid marks the melting point of a substance at each pressure.

51. Intermolecular Forces © 2012 Pearson Education, Inc. Phase Diagrams Below the triple point the substance cannot exist in the liquid state. Along the solid–gas line those two phases are in equilibrium; the sublimation point at each pressure is along this line.

52. Intermolecular Forces © 2012 Pearson Education, Inc. Phase Diagram of Water Note the high critical temperature and critical pressure. –These are due to the strong van der Waals forces between water molecules.

53. Intermolecular Forces © 2012 Pearson Education, Inc. Phase Diagram of Water The slope of the solid– liquid line is negative. –This means that as the pressure is increased at a temperature just below the melting point, water goes from a solid to a liquid.

54. Intermolecular Forces © 2012 Pearson Education, Inc. Phase Diagram of Carbon Dioxide Carbon dioxide cannot exist in the liquid state at pressures below 5.11 atm; CO 2 sublimes at normal pressures.

55. Intermolecular Forces © 2012 Pearson Education, Inc. Sample Exercise 11.5 Interpreting a Phase Diagram Solution Analyze We are asked to identify key features of the phase diagram and to use it to deduce what phase changes occur when specific pressure and temperature changes take place. Plan We must identify the triple and critical points on the diagram and also identify which phase exists at specific temperatures and pressures. Solve (a) The critical point is the point where the liquid, gaseous, and supercritical fluid phases coexist. It is marked point 3 in the phase diagram and located at approximately  80  C and 50 atm. (b) The triple point is the point where the solid, liquid, and gaseous phases coexist. It is marked point 1 in the phase diagram and located at approximately  180  C and 0.1 atm. Use the phase diagram for methane, CH 4, shown in Figure 11.30 to answer the following questions. (a) What are the approximate temperature and pressure of the critical point? (b) What are the approximate temperature and pressure of the triple point? (c) Is methane a solid, liquid, or gas at 1 atm and 0  C? (d) If solid methane at 1 atm is heated while the pressure is held constant, will it melt or sublime? (e) If methane at 1 atm and 0  C is compressed until a phase change occurs, in which state is the methane when the compression is complete?

56. Intermolecular Forces © 2012 Pearson Education, Inc. Sample Exercise 11.5 Interpreting a Phase Diagram Continued (c) The intersection of 0  C and 1 atm is marked point 2 in the phase diagram. It is well within the gaseous region of the phase diagram. (d) If we start in the solid region at P = 1 atm and move horizontally (this means we hold the pressure constant), we cross first into the liquid region, at T   180  C, and then into the gaseous region, at T   160  C. Therefore, solid methane melts when the pressure is 1 atm. (In order for methane to sublime, the pressure must be below the triple point pressure.) (e) Moving vertically up from point 2,which is 1 atm and 0  C, the first phase change we come to is from gas to supercritical fluid. This phase change happens when we exceed the critical pressure (~50 atm). Check The pressure and temperature at the critical point are higher than those at the triple point, which is expected. Methane is the principal component of natural gas. So it seems reasonable that it exists as a gas at 1 atm and 0  C. Practice Exercise Use the phase diagram of methane to answer the following questions. (a) What is the normal boiling point of methane? (b) Over what pressure range does solid methane sublime? (c) Liquid methane does not exist above what temperature? Answer: (a)  162  C; (b) It sublimes whenever the pressure is less than 0.1 atm; (c) The highest temperature at which a liquid can exist is defined by the critical temperature. So we do not expect to find liquid methane when the temperature is higher than  80  C.

57. Intermolecular Forces © 2012 Pearson Education, Inc. Liquid Crystals Some substances do not go directly from the solid state to the liquid state. In this intermediate state, liquid crystals have some traits of solids and some of liquids.

58. Intermolecular Forces © 2012 Pearson Education, Inc. Liquid Crystals Unlike liquids, molecules in liquid crystals have some degree of order.

59. Intermolecular Forces © 2012 Pearson Education, Inc. Liquid Crystals In nematic liquid crystals, molecules are only ordered in one dimension, along the long axis.

60. Intermolecular Forces © 2012 Pearson Education, Inc. Liquid Crystals In smectic liquid crystals, molecules are ordered in two dimensions, along the long axis and in layers.

61. Intermolecular Forces © 2012 Pearson Education, Inc. Liquid Crystals In cholesteryl liquid crystals, nematic-like crystals are layered at angles to each other.

62. Intermolecular Forces © 2012 Pearson Education, Inc. Liquid Crystals These crystals can exhibit color changes with changes in temperature.

63. Intermolecular Forces © 2012 Pearson Education, Inc. Sample Exercise 11.6 Properties of Liquid Crystals Solution Analyze We have three molecules with different structures, and we are asked to determine which one is most likely to be a liquid crystalline substance. Plan We need to identify all structural features that might induce liquid crystalline behavior. Solve Molecule (i) is not likely to be liquid crystalline because the absence of double and/or triple bonds make this molecule flexible rather than rigid. Molecule (iii) is ionic and the generally high melting points of ionic materials make it unlikely that this substance is liquid crystalline. Molecule (ii) possesses the characteristic long axis and the kinds of structural features often seen in liquid crystals: The molecule has a rodlike shape, the double bonds and benzene rings provide rigidity, and the polar COOCH 3 group creates a dipole moment. Which of these substances is most likely to exhibit liquid crystalline behavior?

64. Intermolecular Forces © 2012 Pearson Education, Inc. Sample Exercise 11.6 Properties of Liquid Crystals Continued Answer: Because rotation can occur about carbon–carbon single bonds, molecules whose backbone consists predominantly of C  C single bonds are too flexible; the molecules tend to coil in random ways and, thus, are not rodlike. Practice Exercise Suggest a reason why decane does not exhibit liquid crystalline behavior. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3