Why is this needle floating?
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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?
2
3
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?
4
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)
+
..
S ..
O
:
O
..
-
dipole-dipole attraction
+
..
S ..
O
:
O
..
-
-
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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10
Boiling points versus molecular mass
100
0
-100
Predict a trend for: NH3, PH3, AsH3, and SbH3
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Predict a trend for: NH3, PH3, AsH3, and SbH3
Boiling Pt (Celcius)
0
-20
0
50
100
NH3
150
SbH3
-40
-60
AsH3
-80
-100
PH3
Molecular Weight (g/mol)
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Boiling Pt (Celcius)
Now let’s look at HF, HCl, HBr, and HI
40
20
HF
0
-20 0
NH3
-40
-60
-80
-100
50
100
SbH3
HI
AsH3
HCl
150
HBr
PH3
Molecular Weight (g/mol)
13
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”.
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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 molecules
are 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.
1.42 Å
1.54 Å
3.35 Å
What type of hybridization is present in each?
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Name
type of solid
Ne
Force(s)
Melting Pt.
(oC)
Boiling Pt.
(oC)
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
n-pentane
iso-pentane
neo-pentane
Hvap=25.8 kJ/mol Hvap=24.7 kJ/mol Hvap=22.8 kJ/mol
All three have the same formula C5H12
Why do they have different enthalpies of vaporization?
London and “Tangling”
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C-C-C-C
C
C
C-C-C
C
iso-pentane
Hvap=24.7 kJ/mol
neo-pentane
Hvap=22.8 kJ/mol
n-pentane
Hvap=25.8 kJ/mol
London and “Tangling”
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Structure effects on boiling points
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Ion-dipole interactions: such as a salt dissolved in water
cation
polar molecule
anion
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Phase changes:
solid  liquid (melting  freezing)
liquid  gas (vaporizing  condensing)
solid  gas (sublimation  deposition)
29
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
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
32
Calculate the enthalpy change upon converting 1 mole of
water from ice at -12oC to steam at 115oC.
solid
-12oC
H1
solid
0oC
+ H2
liquid
100oC
liquid
0o C
+
H3
+
gas
100oC
H4
+
gas
115oC
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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Calculate the enthalpy change upon converting 1 mole of
water from ice at -12oC to steam at 115oC.
solid
-12oC
H1
solid
0oC
+ H2
liquid
100oC
liquid
0o C
+
H3
+
H4
gas
100oc
+
gas
115oc
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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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 a
liquid boils under one atm of pressure.
pressure = 1 atm
vapor pressure = 1 atm
liquid
BOILING
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PHASE DIAGRAMS: (Temperature vs. Pressure)
gas and liquid are
indistinguishable.
critical temperature
and critical pressure
(all 3 phases exists here)
38
H2O
CO2
note slope with pressure
note slope with pressure
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Crystal Structures:
40
unit cells:
contains 1 atom
contains 2 atoms
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