CHAPTER 11 INTERMOLECULAR FORCES

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Slide 1:

1 Why is this needle floating?

Slide 2:

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

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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: the intermolecular attractive forces are strong enough to lock molecules in place (high order). Are they temperature dependent?

Slide 5:

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) + + - -

Slide 6:

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

Slide 7:

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

Slide 8:

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

Slide 9:

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).

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

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

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

Slide 14:

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

Slide 15:

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. little Big and “squashy”

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

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Slide 18:

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

Slide 19:

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?

Slide 20:

20 Metallic Bonding: “sea of e-’s”

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

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

Slide 25:

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

Slide 26:

26 Structure effects on boiling points

Slide 27:

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

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Slide 29:

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

Slide 30:

30 Energy changes accompanying phase changes

Slide 31:

31 Heating curve for 1 gram of water

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

Slide 33:

33 Calculate the enthalpy change upon converting 1 mole of water from ice at -12oC to steam at 115oC. solid -12oC solid 0oC liquid 0oC liquid 100oC gas 100oC gas 115oC 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 Hvap=2260 J/g

Slide 34:

34 Calculate the enthalpy change upon converting 1 mole of water from ice at -12oC to steam at 115oC. solid -12oC solid 0oC liquid 0oC liquid 100oC gas 100oc gas 115oc 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

Slide 35:

35 Vapor pressure

Slide 36:

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

Slide 37:

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

Slide 38:

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