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Ch 10. Particle Forces. 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. Particle Forces Affect. Solubility Vapor Pressures Freezing Points - PowerPoint PPT Presentation
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Ch 10 Particle Forces
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Page 1: Ch 10

Ch 10Particle Forces

Page 2: Ch 10

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.

Page 3: Ch 10

Particle Forces Affect• Solubility

• Vapor Pressures

• Freezing Points

• Boiling Points

Page 4: Ch 10

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)

Page 5: Ch 10

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.

E (Q1Q2)d

Page 6: Ch 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.

Page 7: Ch 10

Size of Ions

Page 8: Ch 10

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)

d)Qk(Q = U 21

Page 9: Ch 10

Comparing Lattice EnergiesLattice Energies of Common

Ionic Compounds

Compound U(kJ/mol)

LiF -1047

LiCl -864

NaCl -790

KCl -720

KBr -691

MgCl2 -2540

MgO -3791

Page 10: Ch 10

PracticeDetermine which salt has the greater lattice

energy.

A. MgO and NaF

B. MgO and MgS

Page 11: Ch 10

Lattice Energy Using Hess’s Law

Page 12: Ch 10

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)ΔHEa = -349 kj/mole

Page 13: Ch 10

Calculating UNa+(g) + e-(g) ---> Na(g) -HIE1

Na(g) ---> Na(s) -Hsub

Cl-(g) ---> Cl(g) + e-(g) -HEA

Cl(g) ---> 1/2Cl2(g) -1/2HBE

Na(s) + 1/2Cl2(g) ---> NaCl(s) Hf

Na+(g) + Cl-(g) ---> NaCl(s) U U = Hf - 1/2HBE - HEA - Hsub - HIE1

Page 14: Ch 10

Lattice energy for NaCl.

Page 15: Ch 10

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.

Page 16: Ch 10

Illustrates of Ion-Dipole Interaction

Page 17: Ch 10

The Solution ProcessBond Breaking Processes

• Break solute particle forces (expanding the solute), endothermic

• Break solvent particle forces (expanding the solvent), endothermic

Page 18: Ch 10

The Solution ProcessAttractive 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

Page 19: Ch 10

The Solution ProcessTheromodynamics

• Enthalpy • Entropy ΔS (Perfect crystal, assumed to be

zero)• Gibbs free energy ΔG

• ΔG = ΔH - T ΔS• ΔG < 0, spontaneous change• ΔG = 0, equilibrium• ΔG > 0, nonspontaneous

Page 20: Ch 10

The Solution ProcessOil 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

Page 21: Ch 10

The Solution ProcessOil 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

Page 22: Ch 10

The Solution ProcessSodium 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.

Page 23: Ch 10

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

Page 24: Ch 10

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

Page 25: Ch 10

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

Page 26: Ch 10

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.

Page 27: Ch 10

Dipole-Dipole ForcesDipole-dipole (Polar molecules)

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%

H Cl H Cl H Clδ–δ–δ– δ+δ+ δ+

Page 28: Ch 10

Slide 28 of 35

Dipole Dipole Interactions

Page 29: Ch 10

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.

Page 30: Ch 10

Slide 30

Hydrogen Bonding in HF(g)

Page 31: Ch 10

Slide 31

Hydrogen Bonding in Water

around a molecule in the solid in the liquid

Page 32: Ch 10

Boiling Points of Binary Hydrides

Page 33: Ch 10

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.

Page 34: Ch 10

Illustrations

Page 35: Ch 10

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.

Page 36: Ch 10

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

Page 37: Ch 10

Slide 37

Instantaneous and Induced Dipoles

Page 38: Ch 10

Molar Mass and Boiling Points of Common Species.

Halogen M(g/mol) Bp(K) Noble Gas M(g/mol) Bp(K)

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 211 211

Molar Mass and Boiling Point

Page 39: Ch 10

Hydrocarbon AlcoholMolecular Formula

Molar Mass

Bp (oC)

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 58.12 -0.5 CH3CH2CH2OH 60.09 97

London vs Hydrogen Bonding

Page 40: Ch 10

The Effect of Shape on Forces

Page 41: Ch 10

Practice Rank the following compound in order of increasing

boiling point. CH3OH, CH3CH2CH2CH3, and CH3CH2OCH3

Page 42: Ch 10

Practice Rank the following compound in order of increasing

boiling point. CH3OH, CH3CH2CH2CH3, and CH3CH2OCH3

CH3OH

CH3CH2CH2CH3

CH3CH2OCH3

MM32.0

58.0

60.0

IM ForcesLondon and H-bonding

London, only

London and Dipole-dipole

Page 43: Ch 10

Practice Rank the following compound in order of increasing

boiling point. CH3OH, CH3CH2CH2CH3, and CH3CH2OCH3

CH3OH

CH3CH2CH2CH3

CH3CH2OCH3

MM32.0

58.0

58.0

IM ForcesLondon and H-bonding

London, only

London and Dipole-dipole

The order is:

CH3CH2CH2CH3 < CH3CH2OCH3< CH3OH

Page 44: Ch 10

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

Page 45: Ch 10

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.

Page 46: Ch 10

Polarity and SolubilityHow 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.

Page 47: Ch 10

Polarity and SolubilityHow 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

Page 48: Ch 10

Polarity and SolubilityHow 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.

Page 49: Ch 10

Practice Rank the following compound in order

of increasing boiling point. CH3OH, CH3CH2CH2CH3, and CH3CH2OCH3

Page 50: Ch 10

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.

Page 51: Ch 10

Henry’s Law ConstantsHenry’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

Page 52: Ch 10

Terms• A hydrophobic (“water-fearing) interaction

repels water and diminishes water solubility.

• A hydrophilic (“water-loving”) interaction attracts water and promotes water solubility.

Page 53: Ch 10

Affects of Intermolecular Forces• Solubility

• Vapor Pressures

• Freezing Points

• Boiling Points

• Surface tension

Page 54: Ch 10

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.

Page 55: Ch 10

Vapor Pressure

Page 56: Ch 10

Vapor Pressure

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

Page 57: Ch 10

Vapor Pressure of SolutionsWhat evaporates faster, sugar water or pure water?

s w s w s w w w w w w w

Page 58: Ch 10

Vapor Pressure of SolutionsWhat evaporates faster, sugar water or pure water?

s w s w s w w w w w w w

w(g)w(g) w(g)

Page 59: Ch 10

Vapor Pressure of SolutionsWhat evaporates faster, sugar water or pure water?

s w s w s w w w w w w w

w(g)w(g) w(g)

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.

Page 60: Ch 10

July 2009 General Chemistry: Chapter 11 Slide 60 of 46

Vapor Pressures of Solutions• Roault, 1880s.

Dissolved solute lowers vapor pressure of solvent.

The partial pressure exerted by solvent vapor above an ideal solution is the product of the mole fraction of solvent in the solution and the vapor pressure of the pure solvent at a given temperature.

PA = A P°

Page 61: Ch 10

Vapor Pressure of Solutions

• Raoult’s Law

Psolution = Xsolvent (Psolvent)

• P - vapor pressure

• X - mole fraction

• Xsolute + Xsolvent = 1

Page 62: Ch 10

Vapor pressure of solutions• What evaporates faster pure water of

sugar water?• What is vapor pressure• Roualts Law Psol = XsolventP°solvent

Equation of a straight line The slope is the P°solvent

The y-intercept is zero

July 2009 General Chemistry: Chapter 11 62

Page 63: Ch 10

For a Solution that Obeys Raoult's Law, a Plot fo Psoln Versus Xsolvent, Give a Straight Line

Page 64: Ch 10

Vapor Pressure of Solvent and Solution

Page 65: Ch 10

An Aqueous Solution and Pure Water in a Closed Environment

Page 66: Ch 10

July 2009 General Chemistry: Chapter 11 Slide 66 of 46

Liquid-Vapor Equilibrium

Page 67: Ch 10

Two Volatile LiquidsPositive deviationIdeal Solution Negative deviation

Positive deviation exists when experimental value is larger than calculated value, weaker solute solvent attraction; more evaporation.

Negative deviation exists when experimental value is smaller than calculated value; stronger solvent solute attraction; less evaporation

Page 68: Ch 10

July 2009 General Chemistry: Chapter 11 Slide 68 of 46

Fractional Distillation

Page 69: Ch 10

July 2009 General Chemistry: Chapter 11 Slide 69 of 46

Fractional Distillation

Page 70: Ch 10

Phase Diagram of WaterNote: Negative

slope

Page 71: Ch 10

Phase Diagram of Water

Page 72: Ch 10

Phase Diagram CO2Note:

Posititve Slope

Page 73: Ch 10

Phase Diagram CO2

Page 74: Ch 10

Special Properties of WaterHydrogen bonding found in water gives it special properties listed below:

• Surface tension

• Capillary Rise

• Viscosity

Page 75: Ch 10

Surface Tension

Page 76: Ch 10

Surface TensionSurface Tension units J/m2

Page 77: Ch 10

Surface TensionAdhesive forces stronger in red aqueous solution

Cohesive forces stronger in mercury liquid

Page 78: Ch 10

Capillary Rise

Page 79: Ch 10

Viscosity

Page 80: Ch 10

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.

Page 81: Ch 10

THE END

Page 82: Ch 10

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.

Page 83: Ch 10

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.

Page 84: Ch 10

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.

Page 85: Ch 10

ChemTour: Molecular Motion

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Students use an interactive graph to explore the relationship between kinetic energy and temperature. Includes Practice Exercises.

Page 86: Ch 10

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.

Page 87: Ch 10

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.

Page 88: Ch 10

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.

Page 89: Ch 10

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.

Page 90: Ch 10

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.

Page 91: Ch 10

 

Solubility of CH4, CH2Cl2, and CCl4Which of the following three compounds is most soluble in water?

A) CH4(g) B) CH2Cl2(λ) C) CCl4(λ)

Page 92: Ch 10

Solubility of CH4, CH2Cl2, and CCl4

Consider the following arguments for each answer and vote again:

A. A gas is inherently easier to dissolve in a liquid than is another liquid, since its density is much lower.

B. The polar molecule CH2Cl2 can form stabilizing dipole-dipole interactions with the water molecules, corresponding to a decrease in ΔH°soln.

C. 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.

Page 93: Ch 10

The End


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