1. Forces Between Ions and Molecules and Colligative Properties
Chapter 10
Forces Between Ions and Molecules and Colligative Properties

2. States of Matter
The state of a pure substance (at a given T & P) is governed by intermolecular (non-bonding) forces

3. At room temperature and pressure, He, Ar, Kr, Xe, and Rn are monatomic gases; H2, N2, O2, F2, and Cl2 are diatomic gases; Hg and Br2 are liquids. All other elements are solids.

4. Bonding Forces

5. Intermolecular Forces
E(kJ/mol)

6. Intermolecular Forces
Ion-ion Interactions
Ion-dipole Interactions
Dipole-dipole Interactions
Hydrogen bonds
Dispersion Forces (London Forces)
Dipole-induced Dipole interactions

7. Induced dipole* – Induced dipole* 0.05 – 2
Intermolecular ENERGY
Interaction Type (kJ/mol)
Ion – Ion – 600
Ion – Dipole – 100
Dipole – Dipole – 10
H-Bonds – 40
Ion – Induced Dipole – 10
Dipole – Induced Dipole* – 10
Induced dipole* – Induced dipole* – 2
*Induced dipole forces also called London Dispersion Forces

8. 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.

9. Substances that possess charges will be attracted or repelled by one-another. The energy of these electrostatic interactions is described by Coulomb’s Law:
E = 2.31E -21 J۰nm (Q1Q2/d)
Where Q represents the charge on the substance (e.g. an ion) and d is the distance separating the two charges.
E is negative when the charges have opposite sign (+/-), and positive otherwise.

10. 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.

11. Size of Ions

12. Lattice Energy
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)

13. Ionic compounds are generally solids at room temperature
Ionic compounds are generally solids at room temperature. They are solids because intermolecular attraction between ions is the strongest type of intermolecular interaction.
The cations and the anions forming an ionic compound often are arranged in well-defined crystal lattice.
Lattice energies are proportional to the Coulombic interaction energies, but also depend on the specific arrangement of the ions.

14. 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

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

16. Determining Lattice Energy Using Hess’s Law

17. Electron Affinity
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)

18. Successive Electron Affinities (EA) may be defined, e.g.
O(g) + e- → O-(g) DHEA1 = kJ
O-(g) + e- → O2-(g) DHEA2= kJ
O(g) + 2e- → O2-(g) Overall DHEA= kJ

19. Calculating U Na(g) ---> Na(s) -Hsub
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

20. Ionization energy of K = 425 kJ/mol
Problem
Calculate the lattice energy of potassium chloride from the following data:
Ionization energy of K = 425 kJ/mol
Electron Affinity of Cl for 1 e- = -349 kJ/mol
Energy to vaporize K = 89 kJ/mol
Cl2 bond energy = 240 kJ/mol
Energy change for the reaction:
K(s) + ½ Cl2(g) → KCl (s) is -438 kJ/mol .

23. 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.

24. Illustrates of Ion-Dipole Interaction

25. 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.

26. Boiling Points of Binary Hydrides

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

28. Induced Dipoles dipole-induced dipole interaction
induced dipole-induced dipole interaction

29. Strength of Dispersion Forces
The strength of dispersion forces depends on the polarizability of the atoms or molecules involved.
Polarizability (units of cm3) 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.

30. 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

31. 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

32. The Effect of Shape on Forces

33. Practice CH3OH CH3CH2CH2CH3 CH3CH2OCH3 CH3CH2CH3
Rank the following compounds in order of increasing boiling point:
CH3OH
CH3CH2CH2CH3
CH3CH2OCH3
CH3CH2CH3

34. 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.

35. Section 9.5: Polarity and Solubility
Solutes tend to dissolve in solvents in which similar intermolecular interaction are formed, i.e. dipole-dipole, ion-dipole, or induced dipole-induced dipole. This phenomena is generally stated as “like dissolves like”.

36. Section 9.5: Polarity and Solubility
Solutes tend to dissolve (or are miscible) in solvents in which similar intermolecular interactions are formed, i.e. dipole-dipole, ion-dipole, or induced dipole-induced dipole. This phenomena is generally stated as “like dissolves like”.
Aqueous solubility of Alcohols: CH3(CH2)nOH
1-propanol
n = 2
Miscible
1-butanol
n = 3
1.1 M
1-pentanol
n = 4
0.30 M
1-hexanol
n = 5
0.056 M

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

38. Phospholipids are essential for forming cell membranes…
The bilayer is a semipermeable barrier which allows transport of only small non-polar molecules

39. 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.

40. From Henry’s Law an increase in the partial pressure of a gas in the system will result in an increase concentration of dissolved gas.

41. Henry’s Law Constants for Gas Solubility in Water at 20oC
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

42. [N2]aq = KH۰PN2
KH = slope

43. KH values decrease with increased temperatures.
O2(g) ↔ O2(aq) DHsoln < 0

44. Arterial Blood contains about 0. 25 g of oxygen per liter at 37°C (98
Arterial Blood contains about 0.25 g of oxygen per liter at 37°C (98.6°F) and standard atmospheric pressure.
What is the Henry’s Law constant for O2 dissolved in arterial blood? Compare this value to KH for O2 (aq)

46. Strengths of Intermolecular Forces Affects:
Solubility
Vapor Pressure
viscosity
surface tension
Freezing Point
Boiling Point

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

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

49. Vapor Pressure

50. Vapor Pressure and Solute Concentration
Raoult’s Law
Psolution = Xsolvent (Psolvent)
P = vapor pressure
X = mole fraction
An Ideal Solution has vapor pressures consistent with Raoult’s Law

51. Problem
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?

52. 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 (or even concentration).

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

54. 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.

55. Phase Diagram for CO2

56. Surface Tension

57. Cohesive and Adhesive Forces Produce a Meniscus

58. 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.

59. 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.

60. Calculating Changes in Boiling Point
Tb = Kbm
Tb is the increase in Bp
Kb is the boiling-point elevation constant
m is the molality

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

62. 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?

63. 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).
g/mol

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

65. 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?

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

67. Values of van’t Hoff Factors

68. 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).

69. 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

70. Osmosis at the Molecular Level

71. 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

72. 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.

73. 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.

74. Click to launch animation PC | Mac
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.

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

76. 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.

77. 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.

78. 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.

79. 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.

80. 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.

81. Clicker Questions Note
Note to instructors: The following review question is also applicable to chapter 17, The Colorful Chemistry of Transition Metals.

82. 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)
Solubility of CH4, CH2Cl2, and CCl4

83. 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
Solubility of CH4, CH2Cl2, and CCl4