Intermolecular Forces: Introduction
Intermolecular Forces
• Forces between separate molecules and dissolved
ions (not bonds)
• “Van der Waals Forces”
• 15% as strong as covalent or ionic bonds
Chapter 11
Intermolecular Forces: Introduction
• Low temperature – strong
• High temperature – kinetic energy of motion
overcomes the IMF
• Boiling point is a good indicator
– Stronger IMF = higher boiling point
– Weaker IMF = lower boiling point
Ion-Ion
Full to Full Charge
Ion-Dipole
Full to Partial Charge
Hydrogen Bonds
Partial to Partial Charge (H
involved)
Dipole-Dipole
Partial to Partial Charge (No H)
London
Non-polar to Non-Polar
Dispersion Forces
Predict what type of IMF would form between:
a.
b.
c.
d.
e.
Br2 and I2
KCl and water
Water and ammonia
Two SO2 molecules
NaCl ions in a crystal
Type 1: Ion-Ion Forces
• Full Charges to full charges
• High melting points (ionic solids)
Ex:
Na
NaCl
Melting Point
~98 oC
~800 oC
1
Type 2: Ion-Dipole Forces
• Full Charges to partial charges
• Very Strong
Type 3: Hydrogen Bonds
• Stronger than dipole-dipole that do not have
hydrogen (no inner electrons, strong + pull)
• Generally involves hydrogen and O, N or F
R-H · · · · O-R
R-H · · · · N-R
R-H · · · · F-R
Miscibility
Miscibility
• “Like dissolves like”
• Substances that can hydrogen
bond dissolve in one another.
2
Glucose and other sugars
Water Beading
Ice
Ice
DNA
3
Boiling Point
DNA
DNA is TWO
molecules
that are
hydrogen
bonded (like a
zipper)
Type 4: Dipole-Dipole
• Slightly weaker IMF
• Involve + and - charges other than those in
hydrogen bonding
Type 5: London Dispersion Forces
• Very weak IMF
• Caused by temporary imbalances in electrons
• Generally
increases with
increasing molar
mass
• H2O unusually
high - H-bonding
Draw Lewis Dot Structures to explain the
following boiling points
MM (amu)
BP (oC)
CH3CH2CH3
44
-42
CH3CHO
44
21
CH3CN
41
82
London Forces: Inorganic Molecules
• More electrons, more chance for temporary dipole
Boiling Point Table
Halogen
Molar
Mass
BP(oC)
Noble
Gas
Atomic
Mass
BP(oC)
F2 (g)
38.0
-188
He
4.0
-268
Cl2 (g)
71.0
-35
Ne
20.2
-246
Br2 (l)
159.8
59
Ar
39.9
-186
I2 (s)
253.8
185
Kr
83.8
-152
4
Explain the differences in boiling point between Cl2
(-35oC) and Ar (-186oC)
London Forces: Organic Compounds
• The longer the carbon chain, the higher the
London Dispersion Forces (the higher the melting
point and boiling point)
• Chainlike molecules greater London Forces than
“bunched up” molecules (branched)
Ex 1
Rank the following compounds in terms of
increasing melting points: NaCl, CF4, CH3OH.
Ex 2
Separate the following compounds by whether they
have dipole-dipole attractions (including Hbonding) or London Forces. Which should have
the highest dipole-dipole attraction? Which
should have the strongest London Force?
Ex 3
Rank the following in order of increasing boiling
point:
BaCl2, H2, CO, HF, Kr
Br2, Ne, HCl, N2, HF
5
Ex 4
Properties of Liquid
Rank the following in order of increasing boiling
point:
N2, KBr, O2, CH3CH2OH, HCN
• Viscosity – resistance of a liquid to flow
• Oil is more viscous than water
– Water has H-bonds (stronger)
– Oil has London forces (weaker, but there are many
more of them, long carbon chain)
Miscibility
40 W oil
10 W oil
Surface Tension
Surface Tension
6
Capillary Action
• Water is attracted to the
glass
• Mercury more attracted to
itself
Heating Curves
1. Changes of state do not have a temperature
change.
1. Melting/Freezing
2. Boiling/Condensing
2. A glass of soda with ice will stay at 0oC until all
of the ice melts.
3. Graph “flattens out” during changes of state
Heating Curves
Steam heats
up
Temperature
(oC)
Boiling
Heating Curve
No phase change is occurring (heating ice, water, or
steam):
q = mCp T
Melting or boiling:
Water
warms up
Melting
q = mLf
or
q = mLv
Ice warms up
Heat (Joules)
Lf and Lv
Heating Curves
Latent Heat - heat for phase changes. No
temperature change.
Lf –latent heat of freezing/melting
Lv –latent heat of boiling/condensing
Temperature
(oC)
Use q = mLv
Boiling
Use q = mLf
Melting
Use q = mCp T
Heat
(calories)
7
Important Values
Substance
Steam
Water
Ice
2.01
4.18
2.09
Heating Curves: Example 1
How much energy must be removed to cool 100.0
grams of water at 20.0oC to make ice at
–10.0oC?
Cp
J/goC
J/goC
J/goC
Latent Heat of fusion (water)
Lf = 334.7 J/g
Latent Heat of vaporization(water) Lv = 2259.4 J/g
Heating Curves: Example 1
Heating Curves: Example 1
1.
2.
Temperature
(oC)
Melting (q=mLf)
Water cools (q=mCp T)
Cooling the water
q = mCp T = (100 g)(4.18 J/goC)(0oC-20oC)
q = 8360 J (8.36 kJ) (ignore the negative sign for now)
Freezing the water
q = mLf = (100.0 g)(334.7 J/g)= 33.47 kJ
3. Cooling the ice down to –10.0oC
q = mCp T = (100 g)(2.09 J/goC)(-10oC-0oC)
q = 2.09 kJ (we will ignore the negative sign for now)
Ice cools (q=mCp T)
8.36 kJ+ 33.47 kJ+ 2.09 kJ= 43.92 kJ
Heat
Ex 2
Ex 3
How much energy is needed to convert 18.0 grams
of ice at -25oC to steam at 125oC?
How much energy must be used to convert 100.0
grams of water from steam at 110.0oC to ice at
-25.0oC ?
ANS: 56.0 kJ
(309 kJ)
8
Vapor Pressure
• Pressure of a gas above a liquid caused by that
liquid
• Temperature – measure of the average kinetic
energy of molecules
• At any given moment, some molecules have
enough energy to escape
• EX: Even cold water will evaporate
Volatility
• Volatile – liquids that evaporate easily
– Acetone
– Often weak intermolecular forces
• Boiling point – point at which the vapor pressure
of a liquid = vapor pressure of the atmosphere
• Normal Boiling Point – vapor pressure = 1 atm
– Steam pressure cookers – Forces water to boil at a
higher temperature
– High Altitude – water boils at a lower temperature
Four Types of Solids
1.
2.
3.
4.
Molecular Solids (single molecules)
Covalent Network Solids (one large molecule)
Ionic Solids
Metallic Solids
1. Molecular Solids
• Held together by IMF
• Ice
• Plastics
9
2. Covalent Network Solids
•
•
•
•
•
Basically one big molecule
Held together by covalent bonds
Diamond
Graphite
Quartz (SiO2)
3. Ionic Solids
• Held together by electrostatic attraction (Ion:Ion)
• Usually crystalline (unit cells)
4. Metallic Solids
• Atoms share electrons very freely
• Positive nuclei in a “sea of electrons”
• Electrons held loosely
– Conducts electricity
– Photoelectric effect
– Malleable and ductile
Rank by boiling point (low to high). Below each,
tell me which IMF is important:
CH3OH
Cl2
N2
CH3Cl
CH3CH2CH2CH3
CH3CH2CH2OH
CH3OCH3
CH3CH2CH3
Draw Lewis Dots for these
10
(100.0g)(2.01 J/gK)(20.0oC)
(100.0g)(2259.4 J/g)
(100.0g)(4.18 J/gK)(100.0oC)
(100.0g)(334.7 J/g)
(100.0g)(2.09 J/gK)(5.0oC)
Water
qcool
qfreeze
(100)(4.18)(10) =
(100)(334.7)
=
=
4.02 kJ
= 225.9 kJ
= 41.8 kJ
= 33.47 kJ
=
1.05 kJ
306.2 kJ
4.18 kJ
33.47 kJ
37.65 kJ
Ammonia
qv = mLv
m =qv/Lv
m = 37.65 kJ/23.35 kJ/mol = 1.61 mol
mass = (1.61 mol)(17.0 g/mol) = 27.4 g
12.a) Distance greater in liquid state
b) More movement, more volume, lower density
14. Overall, net forces are attractive
16.a) CH3OH has h-bonding, CH3SH does not
b) Xe is heavier, greater London Forces
c) Cl2 more polarizable than Kr
d) Acetone has dipole-dipole forces
18. a) True
b) False c) False d) True
20. a) Br2 b) C5H11SH
c) CH3CH2CH2Cl
2. a) H-bonding, (b)London (c) Ion-dipole (d)
dipole-dipole. Ion-dipole and h-bonding are
stronger
10. a)
Solids = attractive forces (IMF) win
Liquids = Balance
Gases = Kinetic energy wins
b) Increasing T increase KE, eventually
overcoming IMF
c) High pressure forces gas molecules clsoe
together and IMF’s can win
22. Propyl alcohol is longer and more polarizable
24. a) HF has hydrogen bonds, HCl dipole/dipole
b) CHBr3 higher molar mass, more dispersion
c) ICl has dipole-dipole, Br2 only dispersion
26.a) Dispersion, C8H18 higher boiling point
b) C3H8(dispersion) CH3OCH3 (dip-dip)
c) HOOH (h-bonding) HSSH (dip-dip)
d) NH2NH2 (h-bonding) CH3CH3 (dispersion)
32. H2NNH2, HOOH, H2O can all h-bond
11
34.a) Exo b) Endo
38. 275 g CCl2F2
40. 10.3 kJ
c) Endo
d) Exo
12