Why is this needle floating?

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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

100

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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