1. READING FOR TUESDAY: Chapter 18 sections 1 – 3 READING FOR TUESDAY: Chapter 18 sections 1 – 3 HOMEWORK – DUE THURSDAY 12/3/15 HOMEWORK – DUE THURSDAY 12/3/15 HW-BW 12.2 CH 12 #’s 12, 13, 17-25 (all), 28-31 (all) HW-BW 12.2 CH 12 #’s 12, 13, 17-25 (all), 28-31 (all) HOMEWORK – DUE TUESDAY 12/8/15 HOMEWORK – DUE TUESDAY 12/8/15 HW-BW 13 CH 13 #’s 9, 11, 12, 65, 67, 70, 71, 73, 75, 77, 84, 92, 93, 99-103 (all), 107, 126 HW-BW 13 CH 13 #’s 9, 11, 12, 65, 67, 70, 71, 73, 75, 77, 84, 92, 93, 99-103 (all), 107, 126 Lab Lab Wednesday/Thursday – EXP 15 continued Wednesday/Thursday – EXP 15 continued Monday/Tuesday – EXP 16 Monday/Tuesday – EXP 16

2. Boiling Point of Water

3. H2OH2O H2SH2S H 2 Se H 2 Te 100 o C -60 o C -41.5 o C -2.2 o C 18.02 34.08 80.98 129.62 Boiling PointMass (g) CH 4 -161 o C16.05 C 7 H 16 100 o C100.23

4. The Molecular Dance Molecules in the liquid are constantly in motion Molecules in the liquid are constantly in motion vibrational, and limited rotational and translational vibrational, and limited rotational and translational The average kinetic energy is proportional to the temperature The average kinetic energy is proportional to the temperature However, some molecules have more kinetic energy than the average, and others have less However, some molecules have more kinetic energy than the average, and others have less

5. If these high energy molecules are at the surface, they may have enough energy to overcome the attractive forces If these high energy molecules are at the surface, they may have enough energy to overcome the attractive forces therefore – the larger the surface area, the faster the rate of evaporation therefore – the larger the surface area, the faster the rate of evaporation This will allow them to escape the liquid and become a vapor This will allow them to escape the liquid and become a vapor Vapor Pressure

6. Some molecules of the vapor will lose energy through molecular collisions and will get captured back into the liquid when they collide with it Some molecules of the vapor will lose energy through molecular collisions and will get captured back into the liquid when they collide with it Also some may stick and gather together to form droplets of liquid Also some may stick and gather together to form droplets of liquid particularly on surrounding surfaces particularly on surrounding surfaces We call this process condensation We call this process condensation Vapor Pressure

7. Evaporation vs. Condensation Vaporization and condensation are opposite processes Vaporization and condensation are opposite processes In an open container, the vapor molecules generally spread out faster than they can condense In an open container, the vapor molecules generally spread out faster than they can condense So rate of vaporization is greater than the rate of condensation, and there is a net loss of liquid So rate of vaporization is greater than the rate of condensation, and there is a net loss of liquid In a closed container, the vapor is not allowed to spread out indefinitely In a closed container, the vapor is not allowed to spread out indefinitely So in a closed container the rates of vaporization and condensation will be equal at some point So in a closed container the rates of vaporization and condensation will be equal at some point

8. Dynamic Equilibrium In a closed container, once the rates of vaporization and condensation are equal, the total amount of vapor and liquid will not change In a closed container, once the rates of vaporization and condensation are equal, the total amount of vapor and liquid will not change Evaporation and condensation are still occurring, but because they are opposite processes, there is no net gain or loss of either vapor or liquid Evaporation and condensation are still occurring, but because they are opposite processes, there is no net gain or loss of either vapor or liquid When two opposite processes reach the same rate so that there is no gain or loss of material, we call it a dynamic equilibrium When two opposite processes reach the same rate so that there is no gain or loss of material, we call it a dynamic equilibrium this does not mean there are equal amounts of vapor and liquid – it means that they are changing by equal amounts this does not mean there are equal amounts of vapor and liquid – it means that they are changing by equal amounts

9. Dynamic Equilibrium

10. Vapor–Liquid Dynamic Equilibrium If the volume of the chamber is increased, it will decrease the pressure of the vapor inside the chamber If the volume of the chamber is increased, it will decrease the pressure of the vapor inside the chamber fewer vapor molecules in a given volume, causing the rate of condensation to slow fewer vapor molecules in a given volume, causing the rate of condensation to slow For a period of time, the rate of vaporization will be faster than the rate of condensation, and the amount of vapor will increase For a period of time, the rate of vaporization will be faster than the rate of condensation, and the amount of vapor will increase Eventually enough vapor accumulates so that the rate of the condensation increases to the point where it is once again as fast as evaporation Eventually enough vapor accumulates so that the rate of the condensation increases to the point where it is once again as fast as evaporation equilibrium is reestablished equilibrium is reestablished the vapor pressure will be the same as it was before the vapor pressure will be the same as it was before

11. Vapor Pressure The pressure exerted by the vapor when it is in dynamic equilibrium with its liquid is called the vapor pressure The pressure exerted by the vapor when it is in dynamic equilibrium with its liquid is called the vapor pressure Think Dalton’s Law of Partial Pressures to account for the pressure of the water vapor when collecting gases by water displacement? Think Dalton’s Law of Partial Pressures to account for the pressure of the water vapor when collecting gases by water displacement? The weaker the attractive forces between the molecules, the more molecules will be in the vapor The weaker the attractive forces between the molecules, the more molecules will be in the vapor Therefore, the weaker the attractive forces, the higher the vapor pressure Therefore, the weaker the attractive forces, the higher the vapor pressure the higher the vapor pressure, the more volatile the liquid the higher the vapor pressure, the more volatile the liquid

12. The weaker the attractive forces between molecules, the less energy they will need to vaporize The weaker the attractive forces between molecules, the less energy they will need to vaporize weaker attractive forces means that more energy will need to be removed from the vapor molecules before they can condense weaker attractive forces means that more energy will need to be removed from the vapor molecules before they can condense Results in more molecules in the vapor phase, and a liquid that evaporates faster – the weaker the attractive forces, the faster the rate of evaporation Results in more molecules in the vapor phase, and a liquid that evaporates faster – the weaker the attractive forces, the faster the rate of evaporation Liquids that evaporate easily are said to be volatile Liquids that evaporate easily are said to be volatile e.g., gasoline, fingernail polish remover e.g., gasoline, fingernail polish remover Liquids that do not evaporate easily are called nonvolatile Liquids that do not evaporate easily are called nonvolatile e.g., motor oil e.g., motor oil The higher the vapor pressure, the more volatile the liquid The higher the vapor pressure, the more volatile the liquid Vapor Pressure

13. Distribution of Thermal Energy Only a small fraction of the molecules in a liquid have enough energy to escape Only a small fraction of the molecules in a liquid have enough energy to escape As the temperature increases, the fraction of the molecules with “escape energy” increases As the temperature increases, the fraction of the molecules with “escape energy” increases The higher the temperature, the faster the rate of evaporation The higher the temperature, the faster the rate of evaporation

14. Energetics of Vaporization When the high energy molecules are lost from the liquid, it lowers the average kinetic energy When the high energy molecules are lost from the liquid, it lowers the average kinetic energy If energy is not drawn back into the liquid, its temperature will decrease – therefore, vaporization is an endothermic process If energy is not drawn back into the liquid, its temperature will decrease – therefore, vaporization is an endothermic process and condensation is an exothermic process and condensation is an exothermic process Vaporization requires input of energy to overcome the attractions between molecules Vaporization requires input of energy to overcome the attractions between molecules

15. Heat of Vaporization The amount of heat energy required to vaporize one mole of the liquid is called the heat of vaporization,  H vap The amount of heat energy required to vaporize one mole of the liquid is called the heat of vaporization,  H vap sometimes called the enthalpy of vaporization sometimes called the enthalpy of vaporization Always endothermic, therefore  H vap is + Always endothermic, therefore  H vap is + Somewhat temperature dependent Somewhat temperature dependent   H condensation = −  H vaporization

16. Practice – Calculate the amount of heat needed to vaporize 90.0 g of C 3 H 7 OH at its boiling point (  H vap = 39.9 kJ/mol)

17. Vapor Pressure vs. Temperature Increasing the temperature increases the number of molecules able to escape the liquid Increasing the temperature increases the number of molecules able to escape the liquid The net result is that as the temperature increases, the vapor pressure increases The net result is that as the temperature increases, the vapor pressure increases Small changes in temperature can make big changes in vapor pressure Small changes in temperature can make big changes in vapor pressure the rate of growth depends on strength of the intermolecular forces the rate of growth depends on strength of the intermolecular forces

18. Vapor Pressure Curves 760 mmHg normal BP 100 °C BP Ethanol at 500 mmHg 68.1°C

19. a)water b)TiCl 4 c)ether d)ethanol e)acetone Practice – Which of the following is the most volatile? a)water b)TiCl 4 c)ether d)ethanol e)acetone

20. a)water b)TiCl 4 c)ether d)ethanol e)acetone Practice – Which of the following has the strongest Intermolecular attractions? a)water b)TiCl 4 c)ether d)ethanol e)acetone

21. Boiling Point When the temperature of a liquid reaches a point where its vapor pressure is the same as the external pressure, vapor bubbles can form anywhere in the liquid When the temperature of a liquid reaches a point where its vapor pressure is the same as the external pressure, vapor bubbles can form anywhere in the liquid This phenomenon is what is called boiling and the temperature at which the vapor pressure = external pressure is the boiling point This phenomenon is what is called boiling and the temperature at which the vapor pressure = external pressure is the boiling point

22. The normal boiling point is the temperature at which the vapor pressure of the liquid = 1 atm The normal boiling point is the temperature at which the vapor pressure of the liquid = 1 atm The lower the external pressure, the lower the boiling point of the liquid The lower the external pressure, the lower the boiling point of the liquid Boiling Point

23. a)water b)TiCl 4 c)ether d)ethanol e)acetone Practice – Which of the following has the highest normal boiling point? a)water b)TiCl 4 c)ether d)ethanol e)acetone

24. Heating Curve of Water

25. Energy put in Temperature Heating Curve of Water

26. SolidLiquid Heating Curve of Water

27. Energy put in Temperature melting melting point Heating Curve of Water

28. Liquid Heating Curve of Water

29. Liquid Heating Curve of Water

30. Energy put in Temperature melting melting point Heating Curve of Water

31. LiquidGas Heating Curve of Water

32. Energy put in Temperature melting boiling boiling point melting point Heating Curve of Water

33. Energy put in melting boiling SLG heat of vaporization:heat of fusion: specific heat water: specific heat steam: specific heat ice:

34. Heating Curve of Water

35. Some… Math If you have 100.0 grams of… Water at 100 o C, what is the energy change if the water changes to 37 o C? Steam at 100 o C, what is the energy change if the it condenses to water at 100 o C?

36. Heating Curve of a Liquid As you heat a liquid, its temperature increases linearly until it reaches the boiling point As you heat a liquid, its temperature increases linearly until it reaches the boiling point q = mass x C s x  T q = mass x C s x  T Once the temperature reaches the boiling point, all the added heat goes into boiling the liquid – the temperature stays constant Once the temperature reaches the boiling point, all the added heat goes into boiling the liquid – the temperature stays constant Once all the liquid has been turned into gas, the temperature can again start to rise Once all the liquid has been turned into gas, the temperature can again start to rise

37. The logarithm of the vapor pressure vs. inverse absolute temperature is a linear function A graph of ln(P vap ) vs. 1/T is a straight line The slope of the line x 8.314 J/mol∙K = DH vap in J/mol Clausius–Clapeyron Equation The graph of vapor pressure vs. temperature is an exponential growth curve The graph of vapor pressure vs. temperature is an exponential growth curve

38. Determine the  H vap of dichloromethane given the vapor pressure vs. temperature data Enter the data into a spreadsheet and calculate the inverse of the absolute temperature and natural log of the vapor pressure Enter the data into a spreadsheet and calculate the inverse of the absolute temperature and natural log of the vapor pressure

39. Graph the inverse of the absolute temperature vs. the natural log of the vapor pressure Graph the inverse of the absolute temperature vs. the natural log of the vapor pressure Determine the  H vap of dichloromethane given the vapor pressure vs. temperature data

40. Add a trendline, making sure the display equation on chart option is checked off Add a trendline, making sure the display equation on chart option is checked off Determine the  H vap of dichloromethane given the vapor pressure vs. temperature data

41. Determine the slope of the line Determine the slope of the line −3776.7 ≈ −3800 K −3776.7 ≈ −3800 K Determine the  H vap of dichloromethane given the vapor pressure vs. temperature data

42. Example 11.4: Determine the  H vap of dichloromethane given the vapor pressure vs. temperature data Use the slope of the line to determine the heat of vaporization Use the slope of the line to determine the heat of vaporization slope ≈ −3800 K, R = 8.314 J/mol∙K slope ≈ −3800 K, R = 8.314 J/mol∙K

43. Clausius–Clapeyron Equation The Clausius-Clapeyron equation can be used with just two measurements of vapor pressure and temperature The Clausius-Clapeyron equation can be used with just two measurements of vapor pressure and temperature Can also be used to predict the vapor pressure if you know the heat of vaporization and the normal boiling point Can also be used to predict the vapor pressure if you know the heat of vaporization and the normal boiling point remember: the vapor pressure at the normal boiling point is 760 torr remember: the vapor pressure at the normal boiling point is 760 torr

44. Practice – Determine the vapor pressure of water at 25  C (normal BP = 100.0  C,  H vap = 40.7 kJ/mol)

45. Melting = Fusion As a solid is heated, its temperature rises and the molecules vibrate more vigorously As a solid is heated, its temperature rises and the molecules vibrate more vigorously Once the temperature reaches the melting point, the molecules have sufficient energy to overcome some of the attractions that hold them in position and the solid melts (or fuses) Once the temperature reaches the melting point, the molecules have sufficient energy to overcome some of the attractions that hold them in position and the solid melts (or fuses) The opposite of melting is freezing The opposite of melting is freezing

46. Heating Curve of a Solid As you heat a solid, its temperature increases linearly until it reaches the melting point As you heat a solid, its temperature increases linearly until it reaches the melting point q = mass x C s x  T q = mass x C s x  T Once the temperature reaches the melting point, all the added heat goes into melting the solid – the temperature stays constant Once the temperature reaches the melting point, all the added heat goes into melting the solid – the temperature stays constant Once all the solid has been turned into liquid, the temperature can again start to rise Once all the solid has been turned into liquid, the temperature can again start to rise ice/water will always have a temperature of 0 °C ice/water will always have a temperature of 0 °C at 1 atm at 1 atm

47. Energetics of Melting When the high energy molecules are lost from the solid, it lowers the average kinetic energy When the high energy molecules are lost from the solid, it lowers the average kinetic energy If energy is not drawn back into the solid its temperature will decrease – therefore, melting is an endothermic process If energy is not drawn back into the solid its temperature will decrease – therefore, melting is an endothermic process and freezing is an exothermic process and freezing is an exothermic process Melting requires input of energy to overcome the attractions between molecules Melting requires input of energy to overcome the attractions between molecules

48. Heat of Fusion The amount of heat energy required to melt one mole of the solid is called the Heat of Fusion,  H fus The amount of heat energy required to melt one mole of the solid is called the Heat of Fusion,  H fus sometimes called the enthalpy of fusion sometimes called the enthalpy of fusion Always endothermic, therefore  H fus is + Always endothermic, therefore  H fus is + Somewhat temperature dependent Somewhat temperature dependent  Generally much less than  H vap   H sublimation =  H fusion +  H vaporization

49. Heats of Fusion and Vaporization

50. Segment 1 Heating 1.00 mole of ice at −25.0 °C up to the melting point, 0.0 °C Heating 1.00 mole of ice at −25.0 °C up to the melting point, 0.0 °C q = mass x C s x  T q = mass x C s x  T mass of 1.00 mole of ice = 18.0 g mass of 1.00 mole of ice = 18.0 g C s = 2.09 J/mol∙°C C s = 2.09 J/mol∙°C

51. Segment 2 Melting 1.00 mole of ice at the melting point, 0.0 °C Melting 1.00 mole of ice at the melting point, 0.0 °C q = n∙  H fus q = n∙  H fus n = 1.00 mole of ice n = 1.00 mole of ice  H fus = 6.02 kJ/mol  H fus = 6.02 kJ/mol

52. Segment 3 Heating 1.00 mole of water at 0.0 °C up to the boiling point, 100.0 °C Heating 1.00 mole of water at 0.0 °C up to the boiling point, 100.0 °C q = mass x C s x  T q = mass x C s x  T mass of 1.00 mole of water = 18.0 g mass of 1.00 mole of water = 18.0 g C s = 2.09 J/mol∙°C C s = 2.09 J/mol∙°C

53. Segment 4 Boiling 1.00 mole of water at the boiling point, 100.0 °C Boiling 1.00 mole of water at the boiling point, 100.0 °C q = n∙  H vap q = n∙  H vap n = 1.00 mole of ice n = 1.00 mole of ice  H fus = 40.7 kJ/mol  H fus = 40.7 kJ/mol

54. Segment 5 Heating 1.00 mole of steam at 100.0 °C up to 125.0 °C Heating 1.00 mole of steam at 100.0 °C up to 125.0 °C q = mass x C s x  T q = mass x C s x  T mass of 1.00 mole of water = 18.0 g mass of 1.00 mole of water = 18.0 g C s = 2.01 J/mol∙°C C s = 2.01 J/mol∙°C

55. Practice – How much heat, in kJ, is needed to raise the temperature of a 12.0 g benzene sample from −10.0 °C to 25.0 °C?

56. Phase Diagrams Phase diagrams describe the different states and state changes that occur at various temperature/pressure conditions Phase diagrams describe the different states and state changes that occur at various temperature/pressure conditions Regions represent states Regions represent states Lines represent state changes Lines represent state changes liquid/gas line is vapor pressure curve liquid/gas line is vapor pressure curve both states exist simultaneously both states exist simultaneously critical point is the furthest point on the vapor pressure curve critical point is the furthest point on the vapor pressure curve Triple point is the temperature/pressure condition where all three states exist simultaneously Triple point is the temperature/pressure condition where all three states exist simultaneously

57. Phase Diagrams Pressure Temperature vaporization condensation critical point triple point Solid Liquid Gas 1 atm normal melting pt. normal boiling pt. Fusion Curve Vapor Pressure Curve Sublimation Curve melting freezing sublimation deposition

58. Phase Diagrams

60. Phase Diagram of Water Temperature Pressure critical point 374.1 °C 217.7 atm triple point Ice Water Steam 1 atm normal boiling pt. 100 °C normal melting pt. 0 °C 0.01 °C 0.006 atm

61. Phase Diagram of CO 2 Pressure Temperature critical point 31.0 °C 72.9 atm triple point Solid Liquid Gas 1 atm -56.7 °C 5.1 atm normal sublimation pt. -78.5 °C

64. 20.0 °C, 72.9 atm liquid −56.7 °C, 5.1 atm solid, liquid, gas 10.0 °C, 1.0 atm gas −78.5 °C, 1.0 atm solid, gas 50.0 °C, 80.0 atm scf Consider the phase diagram of CO 2 shown. What phase(s) is/are present at each of the following conditions? 20.0 °C, 72.9 atm −56.7 °C, 5.1 atm 10.0 °C, 1.0 atm −78.5 °C, 1.0 atm 50.0 °C, 80.0 atm

65. Phase Diagram