CHAPTER 11 INTERMOLECULAR FORCES

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1 Why is this needle floating?

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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?

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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) + + - -

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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?

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

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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

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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

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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

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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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17

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

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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?

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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?

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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

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26 Structure effects on boiling points

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

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28

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

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30 Energy changes accompanying phase changes

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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

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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

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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

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35 Vapor pressure

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

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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

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

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39 H2O CO2 note slope with pressure note slope with pressure

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40 Crystal Structures:

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41 unit cells: contains 1 atom contains 2 atoms

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