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Intermolecular Forces: Introduction
• Forces between separate molecules and dissolved ions (not bonds)
• “Van der Waals Forces”
• 15% as strong as covalent or ionic bonds
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 Dispersion Forces
Non-polar to Non-Polar
Predict what type of IMF would form between:
a. Br2 and I2
b.KCl and water
c. Water and ammonia
d.Two SO2 molecules
e. NaCl ions in a crystal
Type 1: Ion-Ion Forces
• Full Charges to full charges
• High melting points (ionic solids)
Ex: Melting Point
Na ~98 oC
NaCl ~800 oC
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
Type 4: Dipole-Dipole
• Slightly weaker IMF
• Involve + and - charges other than those in hydrogen 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 dipoleBoiling 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
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 2
Separate the following compounds by whether they have dipole-dipole attractions (including H-bonding) or London Forces. Which should have the highest dipole-dipole attraction? Which should have the strongest London Force?
Br2, Ne, HCl, N2, HF
Properties of Liquid
• 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)
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
Heat (Joules)
Temperature (oC) Boiling
Melting
Ice warms up
Water warms up
Steam heats up
Heating Curve
No phase change is occurring (heating ice, water, or steam):
q = mCpT
Melting or boiling:
q = mLf or q = mLv
Lf and Lv
Latent Heat - heat for phase changes. No temperature change.
Lf –latent heat of freezing/melting
Lv –latent heat of boiling/condensing
Heating Curves
Heat (calories)
Temperature (oC) Boiling
Melting
Use q = mCpT
Use q = mLf
Use q = mLv
Important Values
Substance Cp
Steam 2.01 J/goC
Water 4.18 J/goC
Ice 2.09 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
How much energy must be removed to cool 100.0 grams of water at 20.0oC to make ice at –10.0oC?
Heating Curves: Example 1
Heat
Temperature (oC)
Melting (q=mLf)
Ice cools (q=mCpT)
Water cools (q=mCpT)
Heating Curves: Example 1
1. Cooling the water
q = mCpT = (100 g)(4.18 J/goC)(0oC-20oC)
q = 8360 J (8.36 kJ) (ignore the negative sign for now)
2. Freezing the water
q = mLf = (100.0 g)(334.7 J/g)= 33.47 kJ
3. Cooling the ice down to –10.0oC
q = mCpT = (100 g)(2.09 J/goC)(-10oC-0oC)
q = 2.09 kJ (we will ignore the negative sign for now)
8.36 kJ+ 33.47 kJ+ 2.09 kJ= 43.92 kJ
Ex 2
How much energy is needed to convert 18.0 grams of ice at -25oC to steam at 125oC?
ANS: 56.0 kJ
Ex 3
How much energy must be used to convert 100.0 grams of water from steam at 110.0oC to ice at -25.0oC ?
(309 kJ)
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. Molecular Solids (single molecules)
2. Covalent Network Solids (one large molecule)
3. Ionic Solids
4. Metallic Solids
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 Draw Lewis Dots for these
CH3OCH3
CH3CH2CH3
(100.0g)(2.01 J/gK)(20.0oC) = 4.02 kJ
(100.0g)(2259.4 J/g) = 225.9 kJ
(100.0g)(4.18 J/gK)(100.0oC) = 41.8 kJ
(100.0g)(334.7 J/g) = 33.47 kJ
(100.0g)(2.09 J/gK)(5.0oC) = 1.05 kJ
306.2 kJ
The heat of vaporization of ammonia is 23.35 kJ/mol. How many grams of ammonia must evaporate to freeze 100.0 grams of water initially at 10.00oC? Assume the ice stays at 0oC when frozen, and is used to cool a cup of too hot tea.
(Ans: 27.4 g)
Water
qcool (100)(4.18)(10) = 4.18 kJ
qfreeze (100)(334.7) = 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
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
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
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