Intermolecular Forces

Slide 1
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Intermolecular Forces
Love & Hate in the Molecular
Realm
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1
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Slide 2
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If I put 2 molecules into a sealed flask,
what could happen?
1.
2.
3.
They ignore each other.
They LOVE each other – they’re attracted
to each other
They HATE each other – they repel each
other
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Slide 3
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REMEMBER:
MOLECULES MOVE!
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(except at 0 K)
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Slide 4
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If they LOVE each other, what would
that look like?
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Initially
Later
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Slide 5
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If they HATE each other, what would
that look like?
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Initially
Later
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Slide 6
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If they IGNORE each other, what
would that look like?
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Initially
Later
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Slide 7
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What determines LOVE or HATE?
The structure of the molecule.
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What is the structure of a molecule?
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H
Br
e-
What’s in the nuclei?
Protons!
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Slide 8
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Molecular structure is all about…
POSITIVE & NEGATIVE CHARGES!
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So Love & Hate is all about…
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Opposites attract, like repel!
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Slide 9
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Types of Intermolecular Forces
1.
2.
3.
4.
5.
London Dispersion forces, aka Van der
Waal’s forces, aka Instantaneous dipoleinduced dipole forces.
Dipole-Dipole interactions
Hydrogen bonding – particularly strong
case of dipole-dipole interaction
Ionic forces
Mixed forces
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Slide 10
London Dispersion forces, aka Van der
Waal’s forces, aka Instantaneous dipoleinduced dipole forces.
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This is NOT the strongest, but it is the
primary intermolecular force.
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All atoms or molecules with electrons have
Van der Waal’s forces – so ALL atoms or
molecules have Van der Waal’s forces
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Slide 11
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Instantaneous dipole-induced dipole
forces
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Induced dipole
Br
δ-
Instantaneous dipole
δ+
Br
Br
δ+
δBr
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The electron cloud is
mobile.
Charge density is
constantly moving
around
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Slide 12
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How Great is THAT!?!?!?
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Induced love
Br
δ-
Instantaneous love
δ+
Br
Br
δ+
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δBr
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Because the induced love is ALWAYS a mirror image of the
instantaneous love, dispersion forces are ALWAYS attractive
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Slide 13
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Dispersion Forces are ALWAYS
ATTRACTIVE
All molecules like each other, at least a little
bit. So all molecules stick together, at
least a little bit.
If they didn’t…
…the universe would be a much more
chaotic place!
Occasional repulsion would have things
flying apart all over the place!
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Slide 14
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Van der Waal’s forces
Van der Waal’s forces get stronger as the
temporary dipole gets stronger.
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The temporary dipole is caused by electron
mobility, so the more electrons the stronger the
Van der Waal’s forces.
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# electrons increases as # protons, so the heavier
the molecule the stronger the Van der Waal’s
forces.
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Slide 15
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Alkanes
Methane – CH4
Ethane – CH3CH3
Propane – CH3CH2CH3
Butane - CH3CH2CH2CH3
Pentane - CH3CH2CH2CH2CH3
Hexane - CH3CH2CH2CH2CH2CH3
Heptane - CH3CH2CH2CH2CH2CH2CH3
Octane - CH3CH2CH2CH2CH2CH2CH2CH3
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Slide 16
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What do they look like?
Here’s propane (C3H8). CH3CH2CH3
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Slide 17
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When you include the electrons…
“Space-filling model”
CH3CH2CH3
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Slide 18
Here’s pentane (C5H12):
CH3CH2CH2CH2CH3
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Slide 19
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Space-filling model of pentane
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Slide 20
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Pentane is just a “longer caterpillar” than
propane.
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That makes it easier to compare these
molecules, they are homologues.
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Slide 21
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What do you know about these
molecules?
Methane – CH4
Ethane – CH3CH3
Propane – CH3CH2CH3
Butane - CH3CH2CH2CH3
Pentane - CH3CH2CH2CH2CH3
Hexane - CH3CH2CH2CH2CH2CH3
Heptane - CH3CH2CH2CH2CH2CH2CH3
Octane - CH3CH2CH2CH2CH2CH2CH2CH3
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Slide 22
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What do you know about these
molecules?
Methane – gas at standard T & P
Ethane – gas at standard T & P
Propane – gas at standard T & P – Liquid under
slight pressure
Butane - gas at standard T & P – Liquid under
slight pressure
Pentane - Liquid
Hexane - Liquid
Heptane - Liquid
Octane - Liquid
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Slide 23
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Solids, Liquids, and Gases
What is the difference between a solid, a liquid, and
a gas microscopically?
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How tightly stuck together the molecules are!!!
Solids are stuck together more than liquids that are
stuck together more than gases
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Slide 24
Sticking together is a function of
TWO things:
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How much you like each other…and how
much you are trying to get away from each
other.
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In the context of molecules, this is a
question of intermolecular forces vs. kinetic
energy.
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Slide 25
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ALL MOLECULES
MOVE!
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(except at 0 K)
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Slide 26
Motion = Kinetic Energy =
Temperature
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Kinetic energy is energy of motion.
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Temperature is a measure of the “mean
kinetic energy of molecules”.
Temperature reflects your desire to escape…
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Slide 27
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Solids, Liquids, and Gases & Heat
What happens when you heat up a solid?
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Eventually it melts – why?
Adding heat adds energy to the molecules,
when they have enough energy they can
escape their attraction to their neighbors!
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Slide 28
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When 2 molecules like each other..
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Slide 29
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Slide 30
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But, they are moving…
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Slide 31
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If they are HOT HOT HOT!
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Slide 32
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If they are very COLLLLDDD
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Slide 33
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Intermolecular forces…
…only depend on distance between the
molecules.
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I can’t change the structure of the molecule.
But the farther apart they are, the smaller
the force they feel.
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[Think gravity and distance from the center
of the earth.]
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Slide 34
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I can add heat…
“Hot” and “Cold” are relative…
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Phase is a balance between the temperature
(kinetic energy) of the molecules that is
trying to separate them and the
intermolecular forces which are trying to
hold them together.
Melting point or boiling point is the kinetic
energy where the balance tips.
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Slide 35
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Van der Waal’s Forces are…
…the first consideration – but not the last!
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But ALL the intermolecular forces are about
CHARGE! (Opposites attract.)
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ALL intermolecular forces are ATTRACTIVE
in the end.
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Slide 36
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Dipole – Dipole Interactions
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Br
Permanent dipole
δ-
Permanent dipole
δ+
H
H
δ+
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δBr
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Slide 37
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Seems REPULSIVE – but it’s really not
Permanent dipole
H
Permanent dipole
δ+
H
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Br
δ+
δ-
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δBr
MOLECULES ARE MOBILE. They always align themselves.
That’s why I say that in the end all intermolecular forces are
attractive.
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Slide 38
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Dipole – Dipole interactions
A molecule with a permanent dipole is called
a “polar molecule”.
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All polar molecules have Dipole-Dipole
interactions in ADDITION TO Van der
Waal’s forces.
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Slide 39
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Dipole – Dipole interactions
Dipole-Dipole interactions are in ADDITION
TO Van der Waal’s forces.
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They are generally weaker and just add on
to VDW forces with ONE EXCEPTION.
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Slide 40
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Hydrogen Bonding – just a special case
of dipole-dipole interactions
Hydrogen bonding is a dipole-dipole
interaction that occurs when hydrogen is
bonded to something very electronegative
like F, O, or N.
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It is just a very strong dipole-dipole
interaction because of the very polar
nature of the H-F, H-O, or H-N bond.
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Slide 41
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Hydrogen Bonding – just a special case
of dipole-dipole interactions
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O
Strongly polar dipole
H
δ--
δ++
Strongly polar dipole
δ++
H
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δ-O
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Slide 42
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Compare H2O to H2S
Which would you expect to have the higher
boiling point?
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H2O has a molar mass of 18 g/mol
H2S has a molar mass of 34 g/mol
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Based on Van der Waal’s forces alone, H2S
should have the higher boiling point.
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Slide 43
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Compare H2O to H2S
The boiling point of water is 373 K.
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The boiling point of H2S is 213 K.
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H2S is a gas at room temperature while
water is a liquid!
No FON, no Hydrogen
bonding
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Slide 44
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Ion – Ion Interactions
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Actual separation of charge
Cl
Na
+
-
Actual separation of charge
+
Na
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Cl
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Slide 45
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Ion-Ion interactions
The strongest possible interaction.
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The complete charge separation makes it a
HUGE dipole-dipole type interaction.
This is why most ionic compounds are solids
at room temperature.
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Slide 46
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Permanent dipole-induced dipole forces
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Induced dipole
Br
δ-
Permanent dipole
δ+
H
Br
δ+
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δBr
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Slide 47
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Dipole – Induced Dipole interactions
This is a special case of a Dipole – Dipole
interaction where there are 2 different
molecules involved and only 1 of them is
polar.
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Generally weaker than a permanent DipoleDipole interaction, it is still IN ADDITION
TO Van der Waal’s forces.
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Slide 48
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All the forces…
1.
2.
3.
4.
Van der Waal’s/Dispersion forces – FIRST
consideration. Weakest for single bond
BUT it is a more global force. Heavier
molecules have bigger VDW forces.
Dipole-Dipole forces – add on to VDW
forces (with ONE exception - #3). If the
molecules have similar mass and shape.
The one with a permanent dipole will
have a higher boiling point.
Hydrogen Bonding TRUMPS VDW
Ionic forces TRUMP EVERYTHING
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Slide 49
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Slide 50
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NaF vs. F2
What do you know about these 2 molecules?
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NaF is an ionic solid
F2 is a gas at room temp
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NaF has a molar mass of 42 g/mol, F2 has a molar
mass of 38 g/mol.
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Slide 51
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Ion-ion interactions are the strongest
Based on Van der Waal’s forces, you’d
expect NaF and F2 to be similar.
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The powerful ionic forces of NaF make it a
solid – trumping the Van der Waal’s
interaction.
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NaF melts at 1266 K and boils at 1968 K
F2 melts at 53 K and boils at 85 K
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Slide 52
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HBr vs. Cl2
What do you know about these 2 molecules?
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HBr is a gas at room temp
Cl2 is a gas at room temp
HBr has a molar mass of 81 g/mol
Cl2 has a molar mass of 71 g/mol
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HBr is polar, Cl2 is non-polar
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Slide 53
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HBr vs Cl2
So HBr is heavier – more van der Waal’s
forces
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HBr is polar – dipole-dipole forces also
So you would think that HBr has the higher
boiling point…and so we go to wikipedia and
find…
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Slide 54
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HBr boils at 207 K, Cl2 boils at 239 K
WTWikipedia?!?!?
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Slide 55
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So, why doesn’t it?
Geometry is also an issue!
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Slide 56
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Geometry
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Br
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H
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Slide 57
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Cl
Cl
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Slide 58
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No separation of charge, no dipole
(including VDW forces)
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Cl
Br
Cl
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H
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Slide 59
This is a warning…your final
warning…
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There are limits to the easy comparisons.
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If you have two structurally similar
molecules, then the heavier one will have
the higher boiling point.
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Slide 60
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Br2 boils at 332 K, Cl2 at 239 K
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Cl
Br
Cl
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Br
Both are non-polar “bar-shaped” molecules
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Slide 61
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Polar molecules are higher…
Permanent dipoles are sort of a bonus.
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Take two similarly shaped molecules with
similar molar masses and the polar one will
have a higher boiling point than the lower
one.
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But if the molar masses are different
enough, the polar nature won’t save you.
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Slide 62
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A few molecules
Molecule
Molar mass
Polar/nonpolar
Boiling point
chloropropane
78.5 g/mol
Weakly Polar
320 K
Hexane
86 g/mol
Non-polar
342 K
Chlorine
70.9 g/mol
Non-polar
239 K
Calcium sulfide
72.1 g/mol
ionic
Melts at 2800 K
Sulfur dioxide
64 g/mol
Polar
263 K
Water
18.02 g/mol
Polar
373 K
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CH3CH2CH2Cl
CH3CH2CH2CH2CH2CH3
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Slide 63
So if you are comparing 2 molecules:
1.
2.
3.
4.
Look for ionic compounds – they have the
strongest forces – trumps EVERYTHING
Look for hydrogen bonding – hydrogen
bonding is the 2nd strongest and will
usually swamp van der Waal’s if the
molecules are SIMILAR size
Van der Waal’s forces – heavier wins
Dipole-dipole forces – sort of a tiebreaker
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Slide 64
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Limits of hydrogen bonding:
Octane – CH3CH2CH2CH2CH2CH2CH2CH3
Molar mass = 114.23 g/mol
Non-polar molecule
Boiling point = 399 K
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Ethanol – CH3CH2-OH
Molar mass =46.07 g/mol
Hydrogen Bonding
Boiling point = 351 K
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Slide 65
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Limits of hydrogen bonding:
Octane – CH3CH2CH2CH2CH2CH2CH2CH3
Molar mass = 114.23 g/mol
Non-polar molecule
Boiling point = 399 K
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Ethanol – CH3CH2-OH
Molar mass =46.07 g/mol
Hydrogen Bonding
Boiling point = 351 K
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Water – H2O
Molar mass = 18.02 g/mol
Hydrogen bonding
Boiling point = 373 K
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Slide 66
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Limits of hydrogen bonding:
Octane – CH3CH2CH2CH2CH2CH2CH2CH3
Molar mass = 114.23 g/mol
Non-polar molecule
Boiling point = 399 K
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Octanol – CH3CH2CH2CH2CH2CH2CH2CH2OH
Molar mass = 130.23 g/mol
Hydrogen bonding
Boiling point = 468 K
Bigger
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Slide 67
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Limits of hydrogen bonding:
Ethanol – CH3CH2-OH
Molar mass =46.07 g/mol
Hydrogen Bonding
Boiling point = 351 K
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Ethane – CH3CH3
Molar mass = 30.07 g/mol
Non-polar
Boiling point = 185 K
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Almost double! The hydrogen bond is a much bigger part of the
smaller molecule
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Slide 68
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Bottom Line
The more similar the molecules are in size
and shape the easier it is to determine the
size of the relative forces.
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If they are very different in size (ethanol vs.
octane) or shape (HBr vs. Cl2) we are just
making educated guesses.
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Slide 69
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In order of importance
1.
2.
3.
4.
Ionic forces (biggest by a lot)
Hydrogen bonding (special case of…)
Dipole-Dipole
Van der Waal’s
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But 3 and 4 are much weaker than 1 and 2.
3 only matters if the molecules are similar
sizes.
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Slide 70
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Slide 71
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Here’s some…
Physical properties that show
“intermolecular forces”:
1. Boiling point
2. Melting point
3. Surface tension
4. Viscosity
5. Capillary action
6. Evaporation
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Slide 72
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Phase changes
Intermolecular Forces are attractions
between molecules.
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Temperature is a measure of kinetic energy.
Boiling Point (or Freezing Point) are
measures of the strength of intermolecular
forces: the higher the temperature, the
more kinetic energy required to separate
the molecules.
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Slide 73
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Not just temperature…
We mentioned TWO things that affected
molecules and their interactions:
1. Energy
2. Space
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Another way of looking at “space” is
pressure.
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Slide 74
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What is “pressure”?
Pressure = Force
Area
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Pressure is squeezing the molecules
together!
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Slide 75
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Phase Changes
You can create a phase change, by changing the
temperature.
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Consider a flask full of steam at 200 C.
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If I start cooling it down, what happens?
It condenses into liquid water. When?
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NOT (necessarily) 100 C.
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Slide 76
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Normal Boiling Point
100 C is the “normal boiling point” of water.
What’s the “normal” for?
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Normal means at standard pressure, 1 atm.
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One way to condense steam is to decrease
the temperature, another way is to
increase the pressure.
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Slide 77
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It’s all about forces!
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Kinetic Energy
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Pressure
Intermolecular
Force
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Slide 78
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Phase Diagrams
A “phase diagram” collects all the P, T and
phase information and displays it in one
simple graph.
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Slide 79
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Phase Diagram for CO2
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Liquid
Pressure
Solid
What do you call this?
Gas
1 atm
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Sublimation!
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-78 C
Temp
79
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Slide 80
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Phase Diagram for H2O
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Liquid
Pressure
1 atm
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Solid
Gas
0 C
Temp
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100 C
80
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Phase Diagrams
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Liquid
1 atm
Pressure
Slide 81
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What do you call this?
Solid
Gas
Triple Point – solid, liquid
and gas coexisting
together!
Temp
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81
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Slide 82
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Phase Diagrams
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Liquid
Pressure
1 atm
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What do you call this?
Solid
Gas
Critical point – No liquid
beyond this – gas of
liquid density
Temp
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Slide 83
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Energy of Phase Changes
How do you define “boiling”?
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Vapor pressure = atmospheric pressure
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What’s vapor pressure?
It’s the pressure exerted by the vapor above
a liquid.
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Slide 84
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As you raise T, you raise Pvap until Pvap = Patm
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Slide 85
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As you raise T, you raise Pvap until Pvap = Patm
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Slide 86
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Remember Ptot = P1 + P2 +
The vapor crowds out the air above the solution
since Ptot must always be Patm
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Slide 87
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Remember Ptot = P1 + P2 +
Ptot must always be Patm. When Pvap = Patm, it’s all
water vapor and WE ARE BOILING!
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Slide 88
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What do you have to do to become “vapor”?
You have to go from a liquid to a gas!
What do you need to do to go from a liquid to a gas?
GAIN ENERGY!
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Slide 89
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Remember, all molecules like each other.
So the difference between a solid, a liquid
and a gas…
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Solid
Liquid
Gas
89
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Slide 90
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…is all relative to the Energy
There are two different energies (or forces).
The attraction between molecules, the
individual energy of the molecules.
___________________________________
___________________________________
___________________________________
Solid
Liquid
Gas
90
___________________________________
___________________________________
___________________________________
Slide 91
___________________________________
Suppose I tie myself to one of you using a
noodle. Could you escape?
___________________________________
___________________________________
___________________________________
91
___________________________________
___________________________________
___________________________________
Slide 92
___________________________________
Of course you could.
You just start walking away and the noodle
breaks.
___________________________________
Suppose I tie myself to you using a piece of
thread?
___________________________________
You may have to walk faster or pull harder
but you can still break away.
___________________________________
92
___________________________________
___________________________________
___________________________________
Slide 93
___________________________________
Suppose I tie myself to you using a piece of
copper wire?
___________________________________
You may have to run or tug or get your
friends to also tug, but you can break the
wire.
___________________________________
___________________________________
93
___________________________________
___________________________________
___________________________________
Slide 94
___________________________________
Same for phases of matter.
They like each other, you want to separate them
you need to overcome the “like”. Easiest way:
heat ‘em up so they are moving faster!
___________________________________
___________________________________
___________________________________
Solid
Liquid
Gas
94
___________________________________
___________________________________
___________________________________
Slide 95
___________________________________
Making a phase change…
Suppose I start with 100 g of ice at -40 C (1 atm)
and start heating it up, what happens?
___________________________________
The ice gets warmer and warmer until…melting
point!
Suppose I am ice at 0 C, do I just spontaneously
melt?
___________________________________
Not exactly. I am warm enough, but I’m still a
solid and my molecules are still “associated” with
each other. I need to get ripped away from my
brothers.
___________________________________
95
___________________________________
___________________________________
___________________________________
Slide 96
___________________________________
___________________________________
O
H
O
H
H
HH
O
HH
O
H H
O
HH
O
H
O
O
O
H H
O
HH
-40 C solid
HH
O
HH
O
H
H
O
H
O
H
H
H
H
HH
O
H H
O
HH
O
H
O
O
O
HH
O
H H
O
HH
H
HH
O
HH
___________________________________
H
O
H
H
___________________________________
0 solid
96
___________________________________
___________________________________
___________________________________
Slide 97
___________________________________
___________________________________
O
H
O
H
HH
O
H
O
H
HH
O
H H
O
HH
O
H H
O
HH
H
H
H
O
HH
O
HH
O
O
O
O
H
H
H
H
H
O
H
O
H
H
H
H
H
O
H
H
O
H
O
___________________________________
H
O
H
___________________________________
H
0 solid
0 liquid
97
___________________________________
___________________________________
___________________________________
Slide 98
___________________________________
At the phase transition temperature…
…you still need energy to make the transition.
Going from solid to liquid, this is called the “heat of
fusion” (Hfus )
___________________________________
___________________________________
Going from liquid to gas, this is called “heat of
vaporization” (Hvap )
Going from solid to gas, this is called the “heat of
sublimation” (Hsub )
___________________________________
98
___________________________________
___________________________________
___________________________________
___________________________________
Phase Diagram for H2O
The “ ” in the H means standard conditions.
___________________________________
Liquid
1 atm
Pressure
Slide 99
___________________________________
Solid
Gas
___________________________________
0 C
Temp
99
___________________________________
___________________________________
___________________________________
Slide 100
___________________________________
Heating/cooling curves
___________________________________
Temp
gas
___________________________________
Liq/gas
liquid
solid
Sol/liq
___________________________________
Heat added
(time, if constantly heating)
100
___________________________________
___________________________________
___________________________________
Slide 101
___________________________________
Heating/cooling curves
___________________________________
Temp
gas
___________________________________
Liq/gas
liquid
Sol/liq
solid
Heat removed
(time, if constantly heating)
___________________________________
101
___________________________________
___________________________________
___________________________________
___________________________________
Could we quantify the energy?
If I have 50.0 g steam at 500 K, how much
energy do I need to remove to get to 373
K?
Q = mcT
___________________________________
500 K
gas
Temp
Slide 102
373 K
Slowing the molecules down.
Liq/gas
___________________________________
liquid
Sol/liq
Heat removed
(time, if constantly heating)
solid
___________________________________
102
___________________________________
___________________________________
___________________________________
Slide 103
___________________________________
Could we quantify the energy?
Once my steam is down to 373 K, how much energy do I need
to remove to turn it into a liquid?
___________________________________
nHvap
gas
Temp
“forming” the
intermolecular forces
___________________________________
Liq/gas
liquid
Sol/liq
solid
___________________________________
Heat removed
(time, if constantly heating)
103
___________________________________
___________________________________
___________________________________
Slide 104
___________________________________
Just follow the curve.
I start here with a gas (gas, gas)
Temp
500 K
Q=mcT
For water:
Csteam = 1.84 J/g c
Cice = 2.09 J/g C
Cwater = 4.18 J/g C
Hfus =6.02 kJ/mol
Hsub =46.72 kJ/mol
Hvap =40.7 kJ/mol
___________________________________
___________________________________
-nHvap
373 K
Q=mcT
-nHfus
Q=mcT
___________________________________
Heat removed
(time, if constantly heating)
104
___________________________________
___________________________________
___________________________________
Slide 105
___________________________________
O
H
___________________________________
H
___________________________________
O
H
___________________________________
H
105
___________________________________
___________________________________
___________________________________
Slide 106
___________________________________
Just follow the curve.
For water:
Csteam = 1.84 J/g c
Cice = 2.09 J/g C
Cwater = 4.18 J/g C
Hfus =6.02 kJ/mol
Hsub =46.72 kJ/mol
Hvap =40.7 kJ/mol
Temp
500 K
373 K
___________________________________
___________________________________
-nHvap
Q=mcT
-nHfus
Q=mcT
Heat removed
(time, if constantly heating)
___________________________________
106
___________________________________
___________________________________
___________________________________
Slide 107
___________________________________
Did I cheat on Temperature?
Why have I mixed K and ºC?
∆T not T
Temp
500 K
___________________________________
A CHANGE in Temperature of 127
K is a change of 127 ºC
373 K
___________________________________
-nHvap
Q=mcT
-nHfus
Q=mcT
Heat removed
(time, if constantly heating)
___________________________________
107
___________________________________
___________________________________
___________________________________
___________________________________
Just follow the curve.
For water:
Csteam = 1.84 J/g c
Cice = 2.09 J/g C
Cwater = 4.18 J/g C
Hfus =6.02 kJ/mol
Hsub =46.72 kJ/mol
Hvap =40.7 kJ/mol
Temp
Slide 108
___________________________________
___________________________________
-nHvap
Q=mcT
Now I’m here…cold…so cold.
-nHfus
Q=mcT
Heat removed
(time, if constantly heating)
___________________________________
108
___________________________________
___________________________________
___________________________________
Slide 109
___________________________________
Cold gas
O
H
___________________________________
H
___________________________________
O
H
___________________________________
H
109
___________________________________
___________________________________
___________________________________
Slide 110
But I don’t want to be a cold gas…I
want to be a liquid
___________________________________
___________________________________
O
H
H
___________________________________
O
H
H
___________________________________
110
___________________________________
___________________________________
___________________________________
___________________________________
Just follow the curve.
For water:
Csteam = 1.84 J/g c
Cice = 2.09 J/g C
Cwater = 4.18 J/g C
Hfus =6.02 kJ/mol
Hsub =46.72 kJ/mol
Hvap =40.7 kJ/mol
Temp
Slide 111
Now, I’m here. Capture!
___________________________________
___________________________________
Q=mcT
-nHfus
Q=mcT
Heat removed
(time, if constantly heating)
___________________________________
111
___________________________________
___________________________________
___________________________________
Slide 112
___________________________________
Just follow the curve.
Temp
For water:
Csteam = 1.84 J/g c
Cice = 2.09 J/g C
Cwater = 4.18 J/g C
Hfus =6.02 kJ/mol
Hsub =46.72 kJ/mol
Hvap =40.7 kJ/mol
___________________________________
___________________________________
Q=mcT
-nHfus
Now, I’m here.
Q=mcT
Heat removed
(time, if constantly heating)
___________________________________
112
___________________________________
___________________________________
___________________________________
Slide 113
___________________________________
YAY! I’M A LIQUID!
___________________________________
O
H
H
___________________________________
O
H
H
___________________________________
113
___________________________________
___________________________________
___________________________________
___________________________________
Just follow the curve.
For water:
Csteam = 1.84 J/g c
Cice = 2.09 J/g C
Cwater = 4.18 J/g C
Hfus =6.02 kJ/mol
Hsub =46.72 kJ/mol
Hvap =40.7 kJ/mol
Temp
Slide 114
___________________________________
373 K
Q=mcT
273 K
___________________________________
-nHfus
Now, I’m here. Instead of a cold
gas, I’m a warm liquid
Heat removed
(time, if constantly heating)
Q=mcT
___________________________________
114
___________________________________
___________________________________
___________________________________
Slide 115
___________________________________
Just follow the curve.
Temp
For water:
Csteam = 1.84 J/g c
Cice = 2.09 J/g C
Cwater = 4.18 J/g C
Hfus =6.02 kJ/mol
Hsub =46.72 kJ/mol
Hvap =40.7 kJ/mol
___________________________________
___________________________________
373 K
-nHfus
273 K
Now, I’m here. Instead of a cold
gas, I’m a warm liquid
Heat removed
(time, if constantly heating)
Q=mcT
___________________________________
115
___________________________________
___________________________________
___________________________________
Slide 116
___________________________________
Just follow the curve.
Temp
For water:
Csteam = 1.84 J/g c
Cice = 2.09 J/g C
Cwater = 4.18 J/g C
Hfus =6.02 kJ/mol
Hsub =46.72 kJ/mol
Hvap =40.7 kJ/mol
___________________________________
___________________________________
373 K
-nHfus
273 K
Now, I’m here. I’m a cold liquid
Heat removed
(time, if constantly heating)
Q=mcT
___________________________________
116
___________________________________
___________________________________
___________________________________
Slide 117
___________________________________
YAY! I’M A LIQUID!
___________________________________
O
H
H
___________________________________
O
H
H
___________________________________
117
___________________________________
___________________________________
___________________________________
Slide 118
___________________________________
I don’t want to be a cold liquid. I
want to be a warm solid!
___________________________________
O
O
O
H
O
H
H
HH
O
H H
O
HH
O
H
O
O
O
HH
O
H H
O
HH
H
H
H
H
H
H
O
HH
O
HH
H
O
O
H
H
O
H
H
O
H
H
O
H
H
H
H
___________________________________
H
O
H
___________________________________
H
0 solid
0 liquid
118
___________________________________
___________________________________
___________________________________
Slide 119
___________________________________
Just follow the curve.
Temp
For water:
Csteam = 1.84 J/g c
Cice = 2.09 J/g C
Cwater = 4.18 J/g C
Hfus =6.02 kJ/mol
Hsub =46.72 kJ/mol
Hvap =40.7 kJ/mol
___________________________________
___________________________________
373 K
-nHfus
273 K
I’m here. I’m a cold liquid
Heat removed
(time, if constantly heating)
I want to
be here.
Warm
Q=mcT solid.
___________________________________
119
___________________________________
___________________________________
___________________________________
___________________________________
Just follow the curve.
For water:
Csteam = 1.84 J/g c
Cice = 2.09 J/g C
Cwater = 4.18 J/g C
Hfus =6.02 kJ/mol
Hsub =46.72 kJ/mol
Hvap =40.7 kJ/mol
Temp
Slide 120
___________________________________
___________________________________
373 K
Now I’m
here.
273 K
Q=mcT down to 0K
Heat removed
(time, if constantly heating)
___________________________________
120
___________________________________
___________________________________
___________________________________
Slide 121
___________________________________
A little problem
I have 50 g of ice at 100 K. How much
energy would I need to add to get steam
at 500 K?
___________________________________
___________________________________
___________________________________
121
___________________________________
___________________________________
___________________________________
Slide 122
___________________________________
Just follow the curve.
For water:
Csteam = 1.84 J/g c
Cice = 2.09 J/g C
Cwater = 4.18 J/g C
Hfus =6.02 kJ/mol
Hsub =46.72 kJ/mol
Hvap =40.7 kJ/mol
___________________________________
Q=mcT
500 K
373 K
Q=mcT
Hfus
___________________________________
Temp
Hvap
273 K
Q=mcT
100 K
___________________________________
Heat removed
(time, if constantly heating)
122
___________________________________
___________________________________
___________________________________
Slide 123
___________________________________
Just follow the curve.
For water:
Csteam = 1.84 J/g c
Cice = 2.09 J/g C
Cwater = 4.18 J/g C
Hfus =6.02 kJ/mol
Hsub =46.72 kJ/mol
Hvap =40.7 kJ/mol
___________________________________
Q=50(1.84)(127)
500 K
(50/18.01)*40700
(50/18.01)6020 J/mol
___________________________________
Temp
373 K
Q=50*4.18*100
273 K
Q=50(2.09)(173)
100 K
Heat removed
(time, if constantly heating)
___________________________________
123
___________________________________
___________________________________
___________________________________
Slide 124
___________________________________
Just follow the curve.
For water:
Csteam = 1.84 J/g c
Cice = 2.09 J/g C
Cwater = 4.18 J/g C
Hfus =6.02 kJ/mol
Hsub =46.72 kJ/mol
Hvap =40.7 kJ/mol
___________________________________
Total Q = 1.81x105 J
1.17x104 J
373 K
Q=2.09x104 J
1.67x104 J
500 K
___________________________________
Temp
1.13x105 J
273 K
1.81x104J
100 K
Heat removed
(time, if constantly heating)
___________________________________
124
___________________________________
___________________________________
___________________________________
Slide 125
___________________________________
Phase Diagram for H2O
For most phase transitions, temperature is more helpful than
pressure…except when a gas is involved.
___________________________________
Liquid
Pressure
1 atm
___________________________________
Solid
Gas
___________________________________
0 C
Temp
125
___________________________________
___________________________________
___________________________________
Slide 126
___________________________________
Hence “vapor pressure”…
…only really applies to sublimation or
boiling.
___________________________________
And unsurprisingly, it depends on
Temperature (how fast the molecules are
moving) and Hvap (how much energy it
takes to separate the molecules and make
them into gases).
___________________________________
___________________________________
126
___________________________________
___________________________________
___________________________________
Slide 127
___________________________________
Vapor Pressure
Vapor pressure depends on temperature. Vapor
pressure also depends on Hvap
___________________________________
Clausius-Clapeyron equation:
___________________________________
Where C is a constant, R is the ideal gas constant.
___________________________________
127
___________________________________
___________________________________
___________________________________
Slide 128
___________________________________
Not completely useful in this form
Clausius-Clapeyron equation:
___________________________________
___________________________________
If I want to calculate Pvap, I need to know
Hvap , C, and T. Except for T, the other
two parameters are specific to each
compound measured. But math (as
ALWAYS!) can Save The Day!!
___________________________________
128
___________________________________
___________________________________
___________________________________
Slide 129
___________________________________
Why did I write it that way?
Clausius-Clapeyron equation:
___________________________________
___________________________________
I could have just written it as:
___________________________________
129
___________________________________
___________________________________
___________________________________
Slide 130
___________________________________
Why did I write it that way?
Clausius-Clapeyron equation:
___________________________________
___________________________________
Looks like:
y = mx+b
___________________________________
130
___________________________________
___________________________________
___________________________________
Slide 131
___________________________________
If I’m doing an experiment
Clausius-Clapeyron equation:
___________________________________
___________________________________
If I plot ln Pvap vs.
I should get a straight
line with a y-intercept of C and a slope of
___________________________________
This is how you would find C and
131
___________________________________
___________________________________
___________________________________
Slide 132
___________________________________
There’s also the short cut
Maybe you don’t want to do the whole
experiment! And maybe someone else has
already determined
(you did the
enthalpy lab!)
___________________________________
___________________________________
Algebra is your BESTEST friend!
___________________________________
132
___________________________________
___________________________________
___________________________________
Slide 133
___________________________________
Common trick
Compare two values
___________________________________
The C and the Hvap depend a little bit on
temperature but not much, so they should be the
same in both equations. So, what do I do?
Simply “compare” the two values by subtracting
them!
___________________________________
___________________________________
133
___________________________________
___________________________________
___________________________________
Slide 134
___________________________________
Common trick
Compare two values
___________________________________
___________________________________
Doing a little algebra…the Cs cancel and we
get…
___________________________________
134
___________________________________
___________________________________
___________________________________
Slide 135
___________________________________
Vapor Pressure
More helpful form – find the Pvap at 2
different temperatures:
___________________________________
___________________________________
This is more helpful for a couple reasons.
First of all…I lost “C”!!! That’s one less
material specific variable to worry about!
___________________________________
135
___________________________________
___________________________________
___________________________________
Slide 136
___________________________________
Vapor Pressure
And then there’s “normal”:
___________________________________
___________________________________
I usually know the “normal boiling point” of
a material…which is?
The boiling point at Patm = 1 atm. Since
boiling occurs when Pvap = Patm, I know
one set of Pvap and T!
___________________________________
136
___________________________________
___________________________________
___________________________________
Slide 137
___________________________________
Sample problem:
What is the vapor pressure of water at
50 C?
___________________________________
I say vapor pressure, you think…
Clausius-Clapeyron!
___________________________________
___________________________________
137
___________________________________
___________________________________
___________________________________
Slide 138
___________________________________
Vapor Pressure
___________________________________
What do I know?
Pvap1 = ?
Pvap2 = ?
Hvap, water = ?
T2 = ?
T1 = ?
R=?
___________________________________
___________________________________
138
___________________________________
___________________________________
___________________________________
Slide 139
___________________________________
Vapor Pressure
___________________________________
What do I know?
Pvap1 = 1 atm
Pvap2 = ?
___________________________________
Hvap, water = 40.7 kJ/mol at boiling point (pg 472, Tro)
= 44.0 kJ/mol at 25 C
T2 = 50 C = 323.15 K
T1 = 100 C = 373.15 K
R = 8.314 J/(mol K)
___________________________________
139
___________________________________
___________________________________
___________________________________
Slide 140
___________________________________
Plugging and chugging time…
___________________________________
What do I know?
Pvap1 = 1 atm
Pvap2 = ?
___________________________________
Hvap, water = 44.0 kJ/mol at 25 C
T2 = 50 C = 323.15 K
T1 = 100 C = 373.15 K
R = 8.314 J/(mol K)
___________________________________
140
___________________________________
___________________________________
___________________________________
Slide 141
___________________________________
Plugging and chugging time…
___________________________________
Whatever you do, DON’T ROUND!
___________________________________
How do I isolate Pvap2?
That’s right ex!
___________________________________
141
___________________________________
___________________________________
___________________________________
Slide 142
___________________________________
Plugging and chugging time…
___________________________________
___________________________________
Does this make sense?
It is less than 1 atm and I’m below the boiling
point!
___________________________________
142
___________________________________
___________________________________
___________________________________
Slide 143
___________________________________
Another little problem
What is the boiling point of water at the top
of Mt. Everest where the average
atmospheric pressure is 0.64 atm?
___________________________________
___________________________________
___________________________________
143
___________________________________
___________________________________
___________________________________
Slide 144
___________________________________
Vapor Pressure
___________________________________
What do I know?
Pvap1 = ?
Pvap2 = ?
Hvap, water = ?
T2 = ?
T1 = ?
R=?
___________________________________
___________________________________
144
___________________________________
___________________________________
___________________________________
Slide 145
___________________________________
Vapor Pressure
___________________________________
What do I know?
Pvap1 = 1 atm
Pvap2 = 0.64 atm
___________________________________
Hvap, water = 44.0 kJ/mol at 25 C
T2 = ?
T1 = 100 C = 373.15 K
R = 8.314 J/(mol K)
___________________________________
145
___________________________________
___________________________________
___________________________________
Slide 146
___________________________________
Plugging and chugging time…
___________________________________
Whatever you do, DON’T ROUND!
___________________________________
ln 1.5625= -5292.278 [0.0026798 – 1/T2]
0.446287 = -5292.278 [0.0026798 – 1/T2]
-0.00008432797 = 0.0026798 – 1/T2
1/T2 = 0.0027642
T2 = 361.77 K = 88.6 C
Does this make sense?
Lower atmospheric pressure, lower boiling point!
___________________________________
146
___________________________________
___________________________________
___________________________________