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Why is this needle floating?. 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?. - PowerPoint PPT Presentation
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Why is this needle floating?
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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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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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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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Types of intermolecular forces (between neutral molecules):
Dipole-dipole forces: (polar molecules)
SO O.. ::
....
:+
--
SO O.. ::
....
:+
--
dipole-dipole attraction
What effect does this attraction have on the boiling point?
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Polar molecules have dipole-dipole attractions for
one another.
+HCl----- +HCl-
dipole-dipole attraction
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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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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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11Predict a trend for: NH3, PH3, AsH3, and SbH3
Boiling points versus molecular mass
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0
-100
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Predict a trend for: NH3, PH3, AsH3, and SbH3
-100
-80
-60
-40
-20
00 50 100 150
Molecular Weight (g/mol)
Boili
ng P
t (Ce
lciu
s) NH3
PH3
AsH3
SbH3
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-100-80-60-40-20
02040
0 50 100 150
Molecular Weight (g/mol)
Boili
ng P
t (Ce
lciu
s)
NH3
PH3
AsH3
SbH3
Now let’s look at HF, HCl, HBr, and HI
HF
HClHBr
HI
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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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polarizability: the ease with which an atom or molecule can be distorted to have an instantaneous dipole. “squashiness”
In general big moleculesare more easily polarized
than little ones.
little Big and“squashy”
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Halogen Boiling Pt (K)
Noble Gas Boiling Pt (K)
F2 85.1 He 4.6
Cl2 238.6 Ne 27.3
Br2 332.0 Ar 87.5
I2 457.6 Kr 120.9
Which one(s) of the above are most polarizable?Hint: look at the relative sizes.
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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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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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Metallic Bonding: “sea of e-’s”
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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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Name type of solid Force(s) Melting Pt.(oC)
Boiling Pt.(oC)
Ne molecular -249 -246
H2S molecular -86 -61
H2O molecular 0 100
Mercury metallic -39 357
W metallic 3410 5660
CsCl ionic 645 1290
MgO ionic 2800 3600
Quartz (SiO2) covalent network 1610 2230
Diamond (C) covalent network 3550 4827
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Pentane isomers: C5H12
iso-pentanen-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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n-pentane
C-C-C-C C
iso-pentane
CC-C-C C
neo-pentane
London and “Tangling”
Hvap=25.8 kJ/mol
Hvap=24.7 kJ/molHvap=22.8 kJ/mol
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Structure effects on boiling points
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Ion-dipole interactions: such as a salt dissolved in water
polar molecule
cation
anion
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Phase changes:
solid liquid (melting freezing)
liquid gas (vaporizing condensing)
solid gas (sublimation deposition)
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Energy changes accompanying phase changes
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Heating curve for 1 gram of water
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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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Calculate the enthalpy change upon converting 1 mole of water from ice at -12oC to steam at 115oC.
solid-12oC
solid0oC
liquid0oC
liquid100oC
gas100oC
gas115oC
H1 + H2 + H3 + H4 + H5 = Htotal
Sp. Ht. + Hfusion + Sp. Ht. + HVaporization + Sp. Ht. = Htotal
Specific Heat of ice = 2.09 J/g•KHfus=334 J/g
Specific Heat of water = 4.184 J/g•KSpecific Ht. Steam = 1.84 J/g•K
Hvap=2260 J/g
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Calculate the enthalpy change upon converting 1 mole of water from ice at -12oC to steam at 115oC.
solid-12oC
solid0oC
liquid0oC
liquid100oC
gas100oc
gas115oc
H1 + H2 + H3 + H4 + H5 = Htotal
Sp. Ht. + Hfusion + Sp. Ht. + HVaporization + Sp. Ht. = Htotal
Specific Heat of ice = 2.09 J/g•KHfus=334 J/g
Specific Heat of water = 4.184 J/g•KSpecific Ht. Steam = 1.84 J/g•K
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Vapor pressure
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VAPOR PRESSURE CURVES
A liquid boils when its vapor pressure =‘s the external pressure.
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normal boiling point is the temperature at which aliquid boils under one atm of pressure.
liquid
pressure = 1 atm
vapor pressure = 1 atm
BOILING
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PHASE DIAGRAMS: (Temperature vs. Pressure)
(all 3 phases exists here)
gas and liquid areindistinguishable.
critical temperatureand critical pressure
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H2O CO2
note slope with pressurenote slope with pressure
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Crystal Structures:
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unit cells:contains 1 atom contains 2 atoms
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