1. Topics in CHM 1046 Intermolecular forces (IMF) Themodynamics
Chemical Kinetics
Chemical Equilibrium
Combination of above:
Unit 11 Intermolecular Forces, Liquids & Solids (T1)
Unit 12 Properties of Solutions (T1)
Unit 13 Thermochemistry (T1)
Unit 14 Chemical Thermodynamics (T2)
Unit 15 Chemical Kinetics (T2)
Unit 16 Chem. Equilibrium Gases & Heterogeneous (T3)
Unit 17.Acid Base Equilibria (T3)
Unit 18 Acid Base Equilibria Buffers & Hydrolysis (T3)
Unit 19 Acid Base Equilibria Titrations (T4)
Unit 20 Aqueous Equilibria Solubility Product (T4)
Unit 21 Electrochemistry (T4)

2. Unit 11 Intermolecular Forces in Liquids and Solids
CHM 1046: General Chemistry and Qualitative Analysis
Unit 11 Intermolecular Forces in Liquids and Solids
Textbook Reference:
Chapter #13
Module #1
Dr. Jorge L. Alonso
Miami-Dade College – Kendall Campus
Miami, FL

3. Molecular Forces: Intra- & Inter-
Intramolecular forces (strong – covalent bond)
Inter- between
Intermolecular forces (weak)
Intra- within
Intermolecular forces (IMF) are strong enough to control physical properties such as boiling and melting points, vapor pressures, and viscosities.

4. Factors affecting IMF
(1) Polarity of the Bond: based on Electronegativity differences between the elements involved in the bond.
=1.9
=0.0
(2) Polarity of the Molecule: based on symmetry of the molecule and how polarity of bonds either cancel or supplement each other.
All have polar bonds
All have polar bonds
Polar Molecule
non-Polar Molecule
non-Polar Molecule
Polar Molecule

5. States of Matter: affected by KE & IMF
Intermolecular Forces (IMF)
Kinetic Energy (Heat, Temperature)
{KE & Phases}
Ideal gas
Condensed phases
PV = nRT

6. Heating Curve: Energy & Phase Changes
Heat of Vaporization
IMF
Heat of Fusion
For H2O:
m.pt. = f.pt = 0OC
b.pt. = c.pt = 100OC
For Methanol:
m.pt. = f.pt = -98OC
b.pt. = c.pt = 65OC
(∆Hv in kJ/mol)

7. Intermolecular Forces
Electronegativity Difference
Non-polar Molecules & Atoms
Polar Molecules
Ions
NaCl He
HCl
H2O O2
Dipole-dipole (& H-bonding)
London Dispersion Forces
Dipole-Induced Dipole
Ion-Dipole
Ionic Bonding
Ions in Solutions
van der Waals Forces (between molecules)
Crystals

8. Liquefaction of Gases: London Dispersion Forces
Boiling (condensation) point
p+ of one atom attracting e- of another atom
Molar Mass {
Air is a mixture of gases, each with a different boiling point. Companies utilize a process known as fractional distillation to separate air gases by using the different boiling points of the gases
Big question: How could a gas as non-polar as He ever turn into a liquid or a solid?

9. London Dispersion Forces
Fritz Wolfgang London
By chance, it does happen that the two electrons in the He atom occasionally wind up on the same side of the atom.
Instantaneous dipole
At that instant, then, the helium atom is polar, with an excess of electrons on the left side and a shortage on the right side.

10. London Dispersion Forces
He atom 2
He atom 1
Another helium nearby, then, would have a dipole induced in it, as the electrons on the left side of helium atom repel the electrons in the cloud on helium atom 1.

11. London Dispersion Forces
Induced
Instantaneous
London dispersion forces are attractions between an instantaneous dipole and an induced dipole.
These forces are present in all molecules, whether they are polar or nonpolar, but they are more noticeable in nonpolar molecules because they are the only IMF betweem such molecules.
The tendency of an electron cloud to distort in this way is called polarizability.

12. Factors Affecting London Forces
What trend does the following data demonstrate about IMF?
IMF
IMF
The strength of dispersion forces tends to increase with increased molecular weight (MW).
Larger atoms have larger electron clouds, which are easier to polarize.
{PhyPropOfHalogens}

13. Distillation of Hydrocarbons: Petroleum Refinery Towers
Factors Affecting London Forces
Distillation of Hydrocarbons: Petroleum Refinery Towers
compounds composed of molecules arranged in a long chain of carbon atoms with hydrogen atoms attached to the carbon chain.
Name (b.pt. C) # C Structural Formula
Methane (-162) 1 CH4
Ethane (-89) 2 CH3CH3
Propane (-42) 3 CH3CH2CH3
Butane (-0.5) 4 CH3CH2CH2CH3
Pentane (36) 5 CH3CH2CH2CH2CH3
Hexane (69) CH3CH2CH2CH2CH2CH3
Heptane (98) CH3CH2CH2CH2CH2CH2CH3
Octane (126) CH3CH2CH2CH2CH2CH2CH2CH3
Nonane (151) CH3 CH2 CH2CH2CH2CH2CH2CH2CH3
Decane (174) 10 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3
IMF & b. pt.

14. Distillation of Hydrocarbons: Petroleum Refinery Towers
Name (b.pt. C) # C Structural Formula
Methane (-162) 1 CH4
Ethane (-89) 2 CH3CH3
Propane (-42) 3 CH3CH2CH3
Butane (-0.5) 4 CH3CH2CH2CH3
Pentane (36) 5 CH3CH2CH2CH2CH3
Hexane (69) CH3CH2CH2CH2CH2CH3
Heptane (98) CH3CH2CH2CH2CH2CH2CH3
Octane (126) CH3CH2CH2CH2CH2CH2CH2CH3
Nonane (151) CH3 CH2 CH2CH2CH2CH2CH2CH2CH3
Decane (174) 10 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3

15. Factors Affecting London Forces
The shape of the molecule affects the strength of dispersion forces: long, skinny molecules (like n-pentane tend to have stronger dispersion forces than short, fat ones (like cyclopentane).
This is due to the increased surface area in n-pentane. C C C C C C
Cyclo-( )

16. Dipole-Dipole Interactions
-polarity
Similar MW
The more polar the molecule, the higher is its boiling point.

17. Which Have a Greater Effect: Dipole-Dipole Interactions or London Dispersion Forces?
Non-polar tail
polar head
Non-polar
tail
polar
head
It depends on the size of the molecules.
If one molecule is much larger than another, dispersion forces will likely determine its physical properties.

18. Hydrogen Bonding Consider the nonpolar series (CH4 to SnH4)
Consider the polar series follows the trend from H2O to H2Te
Increase MW
Increase MW
How would you expect the intermolecular forces to affect the boiling point of the substances in each series?
(increase in MW increase in IMF increase in b.pt.)

19. How Do We Explain This?
The nonpolar series (SnH4 to CH4) follow the expected trend.
The polar series follows the trend from H2Te through H2S,
but water is quite an anomaly.
b.pt.

20. Hydrogen Bonding
The dipole-dipole interactions experienced when H is bonded to N, O, or F are unusually strong.
We call these interactions hydrogen bonds.
{H-bonds}

21. Hydrogen Bonding in DNA
{H-bonding DNA}

23. Factors Determining the State of Matter
CO2
Determining Factors : (1) Polarity (based on Electronegativity) (2) Temperature (Kinetic Energy, heat) (3) Pressure (affects gasses only)
CH3OH
H2O
CO2 (amorphous)
NaCl (crystal)

24. 2001

26. Intermolecular Forces: Adhesion or Cohesion?
Meniscus
Adhesion is attraction between molecules of different substances, i.e., water attracting glass.
Cohesion is the intermolecular attraction between identical molecules, i.e., water attracting water.

27. Liquids Physical Properties Affected by Intermolecular Forces
Viscosity
Surface Tension
Capillary Action
Heat of Vaporization (Energy)
Heat of Fusion (Energy)
Vapor Pressure

28. (1) Viscosity: resistance of a liquid to flow
It is related to the ease with which molecules can move past each other.
Viscosity increases with stronger cohesive intermolecular forces and decreases with higher temperature.
Viscosity generally increases with increased molecular weight (London Forces).
{*Viscosity}

29. (2) Surface Tension {SurfTens}
Results from the net inward cohesive force experienced by the molecules on the surface of a liquid.

30. (3) Capillary Action
{CapAct}
Movement of water thru small capillary tubes (caused by adhesive properties of water & the material in the tube).

31. (4 & 5) Heats of Vaporization & Fusion
Heat of Vaporization
(∆Hv in kJ/mol)
Heat of Fusion
(∆Hf in kJ/mol)
(∆Hv in kJ/mol) 31

32. IMF & Heats of Vaporization/Fusion
Heat of Fusion: Energy required to change a solid at its melting point to a liquid.
(∆Hf in kJ/mol)
IMF
Heat of Vaporization: Energy required to change a liquid at its boiling point to a gas.
(∆Hv in kJ/mol)

33. Energy & Phase Changes
The heat added to the system at the melting and boiling points goes into pulling the molecules farther apart from each other.
{ChangeOfState}
The temperature of the substance does not rise during the phase change.
∆Hv = 41 kJ/mol
How much energy does it take to boil 9g of 100°C?
∆Hf = 6 kJ/mol

34. How much energy does it take to boil 9 g of water @ 100°C?
∆Hv = 41 kJ/mol
= 20.5 kJ
? kJ = 9g H2O
How much energy does it take to melt 9 g of ice 0°C?
∆Hf = 6 kJ/mol
= 3 kJ
? kJ = 9g H2O

35. Heat Capacity: 1. the amount of heat a certain amount of a substance absorbs without changing state
q = m c ∆T
Heat Capacity 35

36. Table of specific heat capacities
Heat Capacity (c)
2. The amount of heat (q) absorbed or released by a particular mass (m) of substance as its temperature changes (T).
Substance
Phase cp
J g−1K−1 
Aluminum
solid
0.897
Copper
0.385
Diamond
0.5091
Gasoline
liquid
2.22
Iron
0.450
Lead
0.127
Oxygen
gas
0.918
Carbon dioxide
0.846
Water
gas (100 °C)
2.080
liquid (25 °C)
4.1813
solid (0 °C)
2.114 ☺
specific heat capacity
change in Temp
heat = (mass)

37. Molar Heat Capacity (cp)
The amount of heat (H) absorbed or released per mole of substance as its temperature changes constant pressure. ☺
molar heat capacity
change in Temp
Enthalpy =

38. Quantifying the Energy of Phase Changes
q = m c ∆T
Heat of vaporization
for H2O: ∆Hv = 41 kJ/mol
cvap = 2.03 J/g0C
q = m c ∆T
cliq = 4.18 J/g0C
cice = 2.09 J/g0C
Heat of fusion for H2O: ∆Hf = 6 kJ/mol
q = m c ∆T

39. Energy & Phase Changes Problems
Given the following information about water: heats of fusion and vaporization, and the specific heat capacities, answer the problem which follows:
cvap = 2.03 J/g0C
∆Hv = 41 kJ/mol
cliq = 4.18 J/g0C
∆Hf = 6 kJ/mol
cice = 2.09 J/g0C
How much heat is required to heat 5 g of water from -4 0C to 110 0C?
(5) From 100  1100C: Q = m c ∆T
(4) ∆Hv = 41 kJ/mol for 5.0 g
(3) From 0  1000C: Q = m c ∆T
(2) ∆Hf = 6 kJ/mol for 5.0 g
(1) From - 4  00C : Q = m c ∆T

40. Energy & Phase Changes Problems
(1) From - 4  00C : q = m c ∆T
cice = 2.09 J/g0C
(5g)(2.09 J/g0C)(4 0C) = 41.8 J
(2) ∆Hf = 6 kJ/mol for 5.0 g
?KJ = 5.0g
= 1.7 kJ
(3) From 0 1000C: q = m c ∆T
cliq = 4.18 J/g0C
(5g)(4.18 J/g0C)(100 0C) = 2,090 J
(4) ∆Hv = 41 kJ/mol for 5.0 g
?KJ = 5.0g
= 11.4 kJ
(5) From 100  1100C: q = m c ∆T
cvap = 2.03 J/g0C
(5g)(2.03 J/g0C)(10 0C) = J
(42J)+(1,700J)+(2,040J)+(11,900J)+(101J)= 15,333 J

41. (6) Vapor Pressure
{*PvapEquil}
{PvapEquil2}
At any temperature, some molecules in a liquid have enough kinetic energy to escape into the gas phase.
The liquid and vapor reach a state of dynamic equilibrium: liquid molecules evaporate and vapor molecules condense at the same rate.
As the temperature rises, the fraction of molecules that have enough energy to escape increases.

42. Vapor Pressure
At lower temperature, some molecules in a liquid have enough kinetic energy to escape into the gas phase.
At higher temperature, the fraction of molecules that have enough energy to escape increases.

43. Vapor Pressure of Water
Vapor Press  (torr)
    
Vapor Press (torr)
-10
2.15 40
55.3
4.58 60
149.4 5
6.54 80
355.1 10
9.21 95
634 11
9.84 96
658 12
10.52 97
682 13
11.23 98
707 14
11.99 99
733 15
12.79
100
760 20
17.54
101
788 25
23.76
110
1074.6 30
31.8
120
1489 37
47.07
200
11659
What kind of relationship is there between temperature and vapor pressure?
Direct or inverse
Linear, exponential (log), parabolic, or gastronomic?
T1-T2 T Pvap
C torr
C 8.33 torr
C torr
Pvap= a e-b/T
Why does water boil at 100 C?
{b.pt. water = 55 C }

44. Vapor Pressure Pvap=ae-b/T
The normal boiling point is the temperature at which its vapor pressure is 760 torr.
The boiling point of a liquid is the temperature at which its vapor pressure equals atmospheric pressure. C
How would the b.pt. of a substance change as the atm. pressure changes?

45. Vapor Pressure and Temperature
Clasius-Claperyon Equation:
Pv = a e-b/T
ln Pv = ln a - b/T
Integrate ☺
Note: The energy units for R (i.e., the units on top) must match those for and T must be in Kelvin.

46. Vapor Pressure and Temperature
Problem: for ethyl alcohol Hvap = 39.3 kJ/mol at its normal boiling point (78.30C). What would be its Pvap (in torr) at 60.00C ?
R = 8.31 x 10-3 kJ/mol

47. Boiling Point of Water? At Patm = 760 torr, b.pt.= 100°C
Mount Everest
Sea level
At Patm= 253 torr, b.pt.= 71°C
Sea level On top of Mount Everest
At what temperature would water boil?

48. Vapor Pressure and Temperature
In Denver the normal atmospheric pressure is 630 torr. At what temperature does water boil in Denver?

49. Vapor Pressure and Temperature

50. Phase Changes at different Pressures
Some Physical
Properties of water:
b. pt. = 1000C
f. pt. = 00C.
∆Hv = 41 kJ/mol
∆Hf = 6 kJ/mol
cvap = 2.03 J/g0C
cliq = 4.18 J/g0C
cice = 2.09 J/g0C
All of these Physical Properties are true only at standard Pressure of 1 atm or 760 mm Hg (torr).

51. Phase Changes at different Pressures
What happens when we consider not only the effect of heat (temperature) but also the effect of pressure?

52. Phase Diagrams: display the state of a substance at various pressures and temperatures and the places where equilibria exist between phases.
1 atm

53. Phase Diagrams
The AB line is the liquid-vapor interface. Each point along this line is the boiling point of the substance at that pressure.
The AD line is the liquid and solid interface. The melting point at each pressure can be found along this line.

54. Phase Diagrams
Critical point (B); above this critical temperature and critical pressure the liquid and vapor are indistinguishable from each other.
{PhaDiag}
Triple Point (A) below this T and P the substance cannot exist in the liquid state. Along the AC line the solid and gas phases are in equilibrium; the sublimation point at each pressure is along this line.

55. Phase Diagram of Water
Note the high critical temperature and critical pressure:
These are due to the strong van der Waals forces between water molecules.

56. Phase Diagram of Water The slope of the solid–liquid line is negative.
This means that as the pressure is increased at a temperature just below the melting point, water goes from a solid to a liquid.

57. Phase Diagram of Carbon Dioxide
Carbon dioxide cannot exist in the liquid state at pressures below 5.11 atm; CO2 sublimes at normal pressures.

58. Phase Diagram of Carbon Dioxide
The low critical temperature and critical pressure for CO2 make supercritical CO2 a good solvent for extracting nonpolar substances (such as caffeine).

59. Solids How many crystals? How many substances? Allotropes
SiO2
SiO2
Allotropes
Crystalline- covalent network
Molecular Solid Sheets
amorphous

60. Solids : Crystalline
We can think of solids as falling into two groups:
Crystalline—particles are in highly ordered arrangement.

61. Solids : Amorphous
Amorphous —no particular order in the arrangement of particles.

62. Types of Intermolecular forces (bonding) in Solids
Ionic
Covalent –Network
Molecular (covalent)
Metallic
(crystalline)
(amorphous)
Which are amorphous or crystalline?

63. Properties of Solid Types
Molecule
Properties
IMF
b.pt. / m.pt.
Hardness
Examples
Ionic
None
Ions held together via electrostatic attractions
(ionic bond)
Very High
Brittle, Cleave on Plane
NaCl,
Ca(NO3)2
Covalent-Network
the whole crystal is a covalently bonded molecule
(covalent bond)
Very Hard, Fractures Erratically
Si,
SiO2 ,
C (diamond)
Molecular
(Polar)
Yes
Covalent compounds existing as solids
Dipole-
Dipole;
H-bonding
Moderate
Soft, Waxy
C12H22O11
(Nonpolar)
Yes, unless Noble Gas
Covalent compoundsor noble gasses
London forces
Moderate to low
Very soft, Waxy
Cl2 , CH4 ,
CO2 , Ar Ne
Metallic
Electrons
Varies
Malleable, Ductile
Cu, Al, Fe, Pt, Au.

64. (1) Ionic Solids Eel =  Q1Q2 d Latice Energy -788 kJ/mol NaCl
Crystal Lattice
Attraction is governed by Coulomb’s law:
(The greater the charge the greater the lattice energy: m.pt. NaCl vs. CaCl2)
Eel = 
Q1Q2 d
Latice Energy -788 kJ/mol

65. Attractions in Ionic Crystals
In ionic crystals, ions pack themselves so as to maximize the attractions and minimize repulsions between the ions.
{ClosePacking}

66. Crystalline Solids
Because of the order in a crystal, we can focus on the repeating pattern of arrangement called the unit cell.
{*UnitCell}
{SimplCubic}
{ClosePacking?}
{Face-Cent1}
{Face-Cent2}

67. (2) Covalent-network Solids

68. (3) Molecular Solids
Graphite is an example of a molecular solid in which atoms are held together with van der Waals forces

69. (4) Metallic Solids
Metals are not covalently bonded, but the attractions between atoms are too strong to be van der Waals forces.
In metals, valence electrons are delocalized throughout the solid.

70. Metallic Bonding: Electron-Sea Model
This explains properties of metals—
Conductivity of heat and electricity
Deformation
Metals can be thought of as cations suspended in “sea” of valence electrons.
Attractions hold electrons near cations, but not so tightly as to impede their flow.

71. Practice Problems

73. 2004 A

75. 2005 A

78. 2005 B (8)

80. 2006 (B)

81. 2006 (B) cont.

82. 2003 A