Why is this needle floating?

1. Why is this needle floating?

2. Intermolecular Forces: (inter = between) between molecules and the temperature (kinetic energy) of the molecules. What determines if a substance is a solid, liquid, or gas?

4. Gases: The average kinetic energy of the gas molecules is much larger than the average energy of the attractions between them. Liquids: the intermolecular attractive forces are strong enough to hold the molecules close together, but without much order. Solids: theintermolecular attractive forces are strong enough to lock molecules in place (high order). Are they temperature dependent?

5. The strengths of intermolecular forces are generally weaker than either ionic or covalent bonds. 16 kJ/mol (to separate molecules) + - + - 431 kJ/mol (to break bond)

6. Types of intermolecular forces (between neutral molecules): Dipole-dipole forces: (polar molecules) .. + S .. : dipole-dipole attraction : O O : .. - - .. + S .. : : O O : .. - - What effect does this attraction have on the boiling point?

7. Polar molecules have dipole-dipole attractions for one another. +HCl----- +HCl- dipole-dipole attraction

8. Types of intermolecular forces (between neutral molecules): Hydrogen bonding: cases of very strong dipole-dipole interaction (bonds involving H-F, H-O, and H-N are most important cases). +H-F- --- +H-F- Hydrogen bonding

9. Hydrogen bonding is a weak to moderate attractive force that exists between a hydrogen atom covalently bonded to a very small and highly electronegative atom and a lone pair of electrons on another small, electronegative atom (F, O, or N).

11. Boiling points versus molecular mass 100 0 -100 Predict a trend for: NH3, PH3, AsH3, and SbH3

12. Predict a trend for: NH3, PH3, AsH3, and SbH3 SbH3 NH3 AsH3 PH3

13. Now let’s look at HF, HCl, HBr, and HI HF SbH3 NH3 HI AsH3 HBr HCl PH3

14. Types of intermolecular forces (between neutral molecules): London dispersion forces: (instantaneous dipole moment) ( also referred to as van der Waal’s forces) attraction - + - + “electrons are shifted to overload one side of an atom or molecule”.

15. polarizability: the ease with which an atom or molecule can be distorted to have an instantaneous dipole. “squashiness” In general big molecules are more easily polarized than little ones. Big and “squashy” little

16. Which one(s) of the above are most polarizable? Hint: look at the relative sizes.

18. Other types of forces holding solids together: ionic: “charged ions stuck together by their charges” There are no individual molecules here.

19. Metallic bonding: “sea of electrons” Copper wire: What keeps the atoms together? Cu atoms an outer shell electron To which nucleus does the electron belong?

20. Metallic Bonding: “sea of e-’s”

21. Covalent Network: (diamonds, quartz) very strong. 1.42 Å 1.54 Å 3.35 Å What type of hybridization is present in each?

24. Pentane isomers: C5H12 neo-pentane n-pentane iso-pentane Hvap=22.8 kJ/mol Hvap=24.7 kJ/mol Hvap=25.8 kJ/mol All three have the same formula C5H12 Why do they have different enthalpies of vaporization? London and “Tangling”

25. C-C-C-C C iso-pentane C C-C-C C neo-pentane Hvap=24.7 kJ/mol Hvap=22.8 kJ/mol n-pentane Hvap=25.8 kJ/mol London and “Tangling”

26. Structure effects on boiling points

27. Ion-dipole interactions: such as a salt dissolved in water cation polar molecule anion

29. Phase changes: solid  liquid (melting  freezing) liquid  gas (vaporizing  condensing) solid  gas (sublimation  deposition)

30. Energy changes accompanying phase changes

31. Heating curve for 1 gram of water

32. Heating curve for 1 gram of water Specific Ht. Steam = 1.84 J/g•K Hvap=2260 J/g Specific Heat of water = 4.184 J/g•K Hfus=334 J/g Specific Heat of ice = 2.09 J/g•K

33. Calculate the enthalpy change upon converting 1 mole of water from ice at -12oC to steam at 115oC. gas 115oC liquid 100oC gas 100oC liquid 0oC solid -12oC solid 0oC H1 + H2 + H3 + H4 + H5 = Htotal Sp. Ht. + Hfusion + Sp. Ht. + HVaporization + Sp. Ht. = Htotal Specific Heat of ice = 2.09 J/g•K Hfus=334 J/g Hvap=2260 J/g Specific Heat of water = 4.184 J/g•K Specific Ht. Steam = 1.84 J/g•K

34. Calculate the enthalpy change upon converting 1 mole of water from ice at -12oC to steam at 115oC. gas 115oc liquid 100oC gas 100oc liquid 0oC solid -12oC solid 0oC H1 + H2 + H3 + H4 + H5 = Htotal Sp. Ht. + Hfusion + Sp. Ht. + HVaporization + Sp. Ht. = Htotal Specific Heat of ice = 2.09 J/g•K Hfus=334 J/g Specific Heat of water = 4.184 J/g•K Specific Ht. Steam = 1.84 J/g•K

35. Vapor pressure

36. VAPOR PRESSURE CURVES A liquid boils when its vapor pressure =‘s the external pressure.

37. normal boiling point is the temperature at which a liquid boils under one atm of pressure. pressure = 1 atm vapor pressure = 1 atm liquid BOILING

38. PHASE DIAGRAMS: (Temperature vs. Pressure) gas and liquid are indistinguishable. critical temperature and critical pressure (all 3 phases exists here)

39. H2O CO2 note slope with pressure note slope with pressure

40. Crystal Structures:

41. unit cells: contains 2 atoms contains 1 atom