1. Chapter 10
Particle Forces

2. States of Matter Solid- Particles moving about a fixed point
Liquid-Particles moving about a moving point
Gas-Particles filling the volume of the container with complete random motions.

3. Particle Forces Affect
Solubility
Vapor Pressures
Freezing Points
Boiling Points 3

4. Particle Forces Intramolecular forces (Relative strength = 100)
Ionic bonding
Covalent bonding
Interparticle forces
Ion-dipole forces
Dipole-dipole (Polar molecules)
(relative Strength = 1)
London Forces (Dispersion forces)( Nonpolar molecules)
(relative strength = 1)
Hydrogen Bonding (Relative strength = 10)

5. Ion-Ion Interactions
Coulomb’s law states that the energy (E) of the interaction between two ions is directly proportional to the product of the charges of the two ions (Q1 and Q2) and inversely proportional to the distance (d) between them.

6. Predicting Forces of Attraction
Coulombs Law indicates the increases in the charges of ions will cause an increase in the force of attraction between a cation and an anion.
Increases in the distance between ions will decrease the force of attraction between them.

7. Size of Ions

8. Lattice Energy M+(g) + X-(g) ---> MX(s)
The lattice energy (U) of an ionic compound is the energy released when one mole of the ionic compound forms from its free ions in the gas phase.
M+(g) + X-(g) ---> MX(s)

9. Comparing Lattice Energies
Lattice Energies of Common Ionic Compounds
Compound
U(kJ/mol)
LiF
-1047
LiCl
-864
NaCl
-790
KCl
-720
KBr
-691
MgCl2
-2540
MgO
-3791

10. Practice Determine which salt has the greater lattice energy.
MgO and NaF
MgO and MgS

11. Lattice Energy Using Hess’s Law

12. Electron Affinity Cl(g) + e-(g) ---> Cl-(g)
Electron affinity is the energy change occurring when one mole of electrons combines with one mole of atoms or ion in the gas phase.
Step 4 in diagram on the last slide.
Cl(g) + e-(g) ---> Cl-(g)
ΔHEa = -349 kj/mole

13. Calculating U Na+(g) + e-(g) ---> Na(g) -HIE1
Na(g) ---> Na(s) -Hsub
Cl-(g) ---> Cl(g) + e-(g) -HEA
Cl(g) ---> 1/2Cl2(g) -1/2HBE
Na(s) + 1/2Cl2(g) ---> NaCl(s) Hf
Na+(g) + Cl-(g) ---> NaCl(s) U
U = Hf - 1/2HBE - HEA - Hsub - HIE1

14. Lattice energy for NaCl.

15. Interactions Involving Polar Molecules
An ion-dipole interaction occurs between an ion and the partial charge of a molecule with a permanent dipole.
The cluster of water molecules that surround an ion in aqueous medium is a sphere of hydration.

16. Illustrates of Ion-Dipole Interaction

17. The Solution Process Bond Breaking Processes
Break solute particle forces (expanding the solute), endothermic
Break solvent particle forces (expanding the solvent), endothermic

18. The Solution Process Attractive Forces
Energy released when solute solvent are attracted, exothermic
Energy is released due to new attractions
Ion dipole if the solute is ionic and the solvent polar.
London-Dipole for nonpolar solute and polar solvent
Dipole-dipole for polar solute and polar solvent

19. The Solution Process Theromodynamics Enthalpy
Entropy (Perfect crystal, assumed to be zero)
Gibbs free energy

20. The Solution Process Oil dissolving in water
London forces holding the oil molecules together are large do to the large surface area of the oil
The hydrogen bonds holding water molecules together are large
The forces of attraction of between nonpolar oil and polar water are weak at best
Thus the overall process is highly endothermic and not allowed thermo chemically

21. The Solution Process Oil dissolving in water
Entropy should be greater than zero
Free energy should be greater than zero, since the process is highly endothermic
Thus the overall process is nonspontaneous

22. The Solution Process Sodium chloride dissolving in water
Large amount of energy is required to break the ionic lattice of the sodium chloride (expand solute)
Large amount of energy is required to separate the water molecules to expand the solvent breaking hydrogen bonds
Formation of the ion dipole forces releases a large amount of energy, strong forces (why?)
The sum of the enthalpies is about +6 kJ (slightly endothermic), which is easily overcome by the entropy of the solution formation.

23. Water as a Solvent
Water most important solvent, important to understand its solvent properties
Most of the unusual solvent properties of water stem from it hydrogen bonding nature
Consider the following ∆S of solution
KCl →75j/K-mole
LiF→-36j/K-mole
CaS→-138 j/K-mole

24. Water as a Solvent We would expect ∆S>0 for all solutions, right?
But two are negative, why?
Obviously, something must be happening for the increased order.
Ion-dipole forces are ordering the water molecules around the ions, thus causing more order in water i.e. less positions for water than in the pure liquid state

25. Water as a Solvent
Smaller ions, have stronger ion dipole forces, thus pulling water closer, therefore less positions
Also, ions with a charge greater than one will attract to water stronger than a one plus charge, thus more order due to less space between particles

26. Dipole-Dipole Interactions
Dipole-dipole interactions are attractive forces between polar molecules.
An example is the interaction between water molecules.
The hydrogen bond is a special class of dipole-dipole interactions due to its strength.

27. Dipole-Dipole Forces H Cl H Cl H Cl Dipole-dipole (Polar molecules) δ+
Alignment of polar molecules to two electrodes
charged + and δ–
Forces compared to ionic/covalent are about 1 in strength compared to a scale of 100, thus 1% δ+ δ– δ+ δ– δ+ δ–
H Cl H Cl H Cl

28. Dipole Dipole Interactions

29. Hydrogen Bonding
Hydrogen bonding a stronger intermolecular force involving hydrogen and usually N, O, F, and sometimes Cl
Stronger that dipole-dipole, about 10 out of 100, or 10
Hydrogen needs to be directly bonded to the heteroatom
Since hydrogen is small it can get close to the heteroatom
Also, the second factor is the great polarity of the bond.

30. Hydrogen Bonding in HF(g)

31. Hydrogen Bonding in Water
Chemistry 140 Fall 2002
Hydrogen Bonding in Water
Solid ice has lower density than liquid water. H-bonding holds the ice in a rigid but open structure.
Maximum density of water at 3.98 C.
around a molecule in the solid in the liquid

32. Boiling Points of Binary Hydrides

33. Interacting Nonpolar Molecules
Dispersion forces (London forces) are intermolecular forces caused by the presence of temporary dipoles in molecules.
A instantaneous dipole (or induced dipole) is a separation of charge produced in an atom or molecule by a momentary uneven distribution of electrons.

34. Illustrations

35. Strength of Dispersion Forces
The strength of dispersion forces depends on the polarizability of the atoms or molecules involved.
Poarizability is a term that describes the relative ease with which an electron cloud is distorted by an external charge.
Larger atoms or molecules are generally more polarizable than small atoms or molecules.

36. London Forces (Dispersion)
Induced dipoles (Instantaneous )
Strength is surface area dependent
More significant in larger molecules
All molecules show dispersion forces
Larger molecules are more polarizable

37. Instantaneous and Induced Dipoles

38. Molar Mass and Boiling Point
Molar Mass and Boiling Points of Common Species.
Halogen
M(g/mol)
Bp(K)
Noble Gas He 2 4 F2 38 85 Ne 20 27
Cl2 71
239 Ar 40 87
Br2
160
332 Kr 84
120 I2
254
457 Xe
131
165 Rn
211

39. London vs Hydrogen Bonding
Hydrocarbon
Alcohol
Molecular Formula
Molar Mass
Bp (oC)
CH4
16.04
-161.5
CH3CH3
30.07
-88
CH3OH
32.04
64.5
CH3CH2CH3
44.09
-42
CH3CH2OH
46.07
78.5
CH3CH(CH)CH3
58.12
-11.7
CH3CH(OH)CH3
60.09 82
CH3CH2CH2CH3
-0.5
CH3CH2CH2OH 97

40. The Effect of Shape on Forces

41. Practice
Rank the following compound in order of increasing boiling point. CH3OH, CH3CH2CH2CH3, and CH3CH2OCH3

42. Practice
Rank the following compound in order of increasing boiling point. CH3OH, CH3CH2CH2CH3, and CH3CH2OCH3 MM
IM Forces
CH3OH
32.0
London and H-bonding
CH3CH2CH2CH3
58.0
London, only
CH3CH2OCH3
60.0
London and Dipole-dipole

43. Practice
Rank the following compound in order of increasing boiling point. CH3OH, CH3CH2CH2CH3, and CH3CH2OCH3 MM
IM Forces
CH3OH
32.0
London and H-bonding
CH3CH2CH2CH3
58.0
London, only
CH3CH2OCH3
58.0
London and Dipole-dipole
The order is:
CH3CH2CH2CH3 <
CH3CH2OCH3<
CH3OH

44. Polarity and Solubility
If two or more liquids are miscible, they form a homogeneous solution when mixed in any proportion.
Ionic materials are more soluble in polar solvents then in nonpolar solvents.
Nonpolar materials are soluble in nonpolar solvents.
Like dissolves like

45. Polarity and Solubility
If two or more liquids are miscible, they form a homogeneous solution when mixed in any proportion.
Ionic materials are more soluble in polar solvents then in nonpolar solvents.
Nonpolar materials are soluble in nonpolar solvents.

46. Polarity and Solubility
How does polarity effect solubility?
The thermodynamic argument, is that the lower the potential energy, the more stable the system. If subtracting the potential energy of the solute from the potential energy of the original solute and solvent is negative (exothermic) then solution is thermodynamically favored.

47. Polarity and Solubility
How does polarity effect solubility?
Non polar solute and solvent: The forces holding these particles together are London Dispersion forces, the weakest of all of the inter-particle forces. The strength of these forces are relative to the surface area if solute and solvent are of similar size, then about the same amount of energy is required to separate solute and solvent particles from each other. And about the same amount of energy is released when solute and solvent are attracted to each other forming a solution. Thus we predict non polar solutes and solvents should dissolve

48. Polarity and Solubility
How does polarity effect solubility?
Non polar solute and polar solvent: Considering solutes and solvents of similar surface area it should be noted that more energy is required to separate the polar solvent molecules from each other, since dipole-dipole interactions are stronger. The only interaction between a nonpolar solute and polar solvent would be London Dispersion forces, so the energy released is much less than required for separating the solvent and solute. Subtracting the potential energy of the products from reactants would give a positive (endothermic) result and the solution would be less stable than the dissolution.

49. Practice
Rank the following compound in order of increasing boiling point. CH3OH, CH3CH2CH2CH3, and CH3CH2OCH3

50. Solubility of Gases in Water
Henry’s Law states that the solubility of a sparingly soluble chemically unreactive gas in a liquid is proportional to the partial pressure of the gas.
Cgas = kHPgas where C is the concentration of the gas, kH is Henry’s Law constant for the gas.

51. Henry’s Law Constants Henry’s Law Constants Gas kH[mol/(L•atm)]
kH[mol/(kg•mmHg)] He
3.5 x 10-4
5.1 x 10-7 O2
1.3 x 10-3
1.9 x 10-6 N2
6.7 x 10-4
9.7 x 10-7
CO2
3.5 x 10-2
5.1 x 10-5

52. Terms
A hydrophobic (“water-fearing) interaction repels water and diminishes water solubility.
A hydrophilic (“water-loving”) interaction attracts water and promotes water solubility.

53. Affects of Intermolecular Forces
Solubility
Vapor Pressures
Freezing Points
Boiling Points
Surface tension

54. Vapor Pressure
Vaporization or evaporation is the transformation of molecules in the liquid phase to the gas phase.
Vapor pressure is the force exerted at a given temperature by a vapor in equilibrium with its liquid phase.

55. Vapor Pressure

56. Vapor Pressure
The normal boiling point of a liquid is the temperature at which its vapor pressure equals 1 atmosphere.

57. Vapor Pressure of Solutions
What evaporates faster, sugar water or pure water?
s w s w s w
w w w w w w

58. Vapor Pressure of Solutions
What evaporates faster, sugar water or pure water?
w(g)
w(g)
w(g)
s w s w s w
w w w w w w

59. Vapor Pressure of Solutions
What evaporates faster, sugar water or pure water?
w(g)
w(g)
w(g)
s w s w s w
w w w w w w
Pure water evaporates faster, since there are more water particles on the surface, thus lowering the average kinetic energy. Evaporation of a solution is inversely proportional to concentration.

60. Vapor Pressure of Solutions
Raoult’s Law
Psolution = Xsolvent (Psolvent)
P - vapor pressure
X - mole fraction
Xsolute + Xsolvent = 1

61. Practice
A solution contains mL of water and mol of ethanol. What is the mole fraction of water and the vapor pressure of the solution at 25oC, if the vapor of pressure of pure water is 23.8 torr?

62. Surface Tension

63. Cohesive and Adhesive Forces Produce a Meniscus

64. Physical State and Phase Transformations
A phase diagram is a graphic representation of the dependence of the stabilities of the physical states of a substance on temperature and pressure.

65. Phase Diagram for Water
Triple Point
Critical Point
Critical Temperature
Critical Pressure
Supercritical Fluid

66. Terms
The triple point defines the temperature and pressure where all three phases of a substance coexist.
The critical point is that specific temperature and pressure at which the liquid and gas phases of a substance have the same density and are indistinguishable for each other.
A supercritical fluid is a substance at conditions above its critical temperature and pressure.

67. Phase Diagram for CO2

68. Terms
Capillary action is the rise of a liquid up a narrow tube as a result of adhesive forces between the liquid and the tube and cohesive forces within the liquid.
Viscosity is a measure of the resistance to flow of a fluid.

69. Colligative Properties of Solutions
Colligative properties of solutions depend on the concentration and not the identity of particles dissolved in the solvent.
Sea water boils at a higher temperature than pure water.

70. Colligative Properties
Colligative property is a physical property that depends on the number of particles present, and not on the nature of the particle. Since evaporation is dependent on the number of solvent particles present on the surface that makes evaporation and vapor pressure colligative properties.

71. Colligative Properties
Is density a colligative property?

72. Colligative Properties
Is density a colligative property?
While density depends on the number of particles in a given area, it is also effected by the weight of the substance, which is a nature thing, so no density is not a colligative property.

73. Calculating Changes in Boiling Point
Tb = Kbm
Tb is the increase in Bp
Kb is the boiling-point elevation constant
m is a new concentration unit called molality

74. Practice
Calculate the molality of a solution containing mol of glucose (C6H12O6) in 1.5 kg of water.

75. Practice
Seawater contains M Cl- at the surface at 25oC. If the density of sea water is g/mL, what is the molality of Cl- in sea water?

76. Practice
Cinnamon owes its flavor and odor to cinnamaldehyde (C9H8O). Determine the boiling-point elevation of a solution of 100 mg of cinnamaldehyde dissolved in 1.00 g of carbon tetrachloride (Kb = 2.34oC/m).

77. Freezing-point Depression
Tf = Kfm
Kf is the freezing-point depression constant and m is the molality.

78. Practice
The freezing point of a solution prepared by dissolving X 102 mg of caffeine in 10.0 g of camphor is 3.07 Celsius degree lower than that of pure camphor (Kf = 39.7oC/m). What is the molar mass of caffeine?

79. The van’t Hoff Factor Tb = iKbm & Tf = iKfm
van’t Hoff factor, i is the number of ions in one formula unit

80. Values of van’t Hoff Factors

81. Practice
CaCl2 is widely used to melt frozen precipitation on sidewalks after a winter storm. Could CaCl2 melt ice at -20oC? Assume that the solubility of CaCl2 at this temperature is 70.0 g/100.0 g of H2O and that the van’t Hoff factor for a saturated solution of CaCl2 is 2.5 (Kf for water is C/m).

82. Osmosis In osmosis, solvent passes through a semipermeable membrane
to balance the concentration of solutes in solution on both sides
of the membrane.
Figure 10.30

83. Osmosis at the Molecular Level

84. Osmotic Pressure
Osmotic pressure () is the pressure that has to be applied across a semipermeable membrane to stop the flow of solvent form the the compartment containing pure solvent or a less concentrated solution towards a more concentrated solution.
 = iMRT where i is the van’t Hoff factor, M is molarity of solute, R is the idea gas constant ( l•atm/(mol•K)), and T is in Kelvin

85. ChemTour: Lattice Energy
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Students learn to apply Coulomb’s law to calculate the exact lattice energies of ionic solids. Includes Practice Exercises.

86. ChemTour: Intermolecular Forces
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This ChemTour explores the different types of intermolecular forces and explains how these affect the boiling point, melting point, solubility, and miscibility of a substance. Includes Practice Exercises.

87. Click to launch animation
ChemTour: Henry’s Law
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Students learn to apply Henry’s law and calculate the concentration of a gas in solution under varying conditions of temperature and pressure. Includes interactive practice exercises.

88. ChemTour: Molecular Motion
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Students use an interactive graph to explore the relationship between kinetic energy and temperature. Includes Practice Exercises.

89. ChemTour: Raoult’s Law
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Students explore the connection between the vapor pressure of a solution and its concentration as a gas above the solution. Includes Practice Exercises.

90. ChemTour: Phase Diagrams
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Students use an interactive phase diagram and animated heating curve to explore how changes in temperature and pressure affect the physical state of a substance.

91. ChemTour: Capillary Action
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In this ChemTour, students learn that certain liquids will be drawn up a surface if the adhesive forces between the liquid on the surface of the tube exceed the cohesive forces between the liquid molecules.

92. ChemTour: Boiling and Freezing Points
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Students learn about colligative properties by exploring the relationship between solute concentration and the temperature at which a solution will undergo phase changes. Interactive exercises invite students to practice calculating the boiling and freezing points of different solutions.

93. ChemTour: Osmotic Pressure
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Students discover how a solute can build up pressure behind a semipermeable membrane. This tutorial also discusses the osmotic pressure equation and the van’t Hoff factor.

94. Solubility of CH4, CH2Cl2, and CCl4
Which of the following three compounds is most soluble in water?
© 2008 W. W. Norton & Company Inc. All rights reserved.
A) CH4(g) B) CH2Cl2(λ) C) CCl4(λ)

95. Solubility of CH4, CH2Cl2, and CCl4
Consider the following arguments for each answer and vote again:
A gas is inherently easier to dissolve in a liquid than is another liquid, since its density is much lower.
The polar molecule CH2Cl2 can form stabilizing dipole-dipole interactions with the water molecules, corresponding to a decrease in ΔH°soln.
The nonpolar molecule CCl4 has the largest molecular mass, and so is most likely to partially disperse into the water, corresponding to an increase in ΔS°soln.
Answer: B

96. The End