logging in or signing up chapter09 Davidino Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINTLite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 3215 Category: Education License: All Rights Reserved Like it (0) Dislike it (0) Added: January 08, 2008 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript ChemistryThe Science in ContextChapter 9Intermolecular Forces and Liquids : Chemistry The Science in Context Chapter 9 Intermolecular Forces and Liquids Slide2: 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. Slide3: Section 9.1 Why is NaCl more soluble in water than CaSO4? Substances can exist in three states: solid, liquid, and gas phases. In which of these phases are there the greatest number of intermolecular interactions? H2O(s) ↔ H2O(l) ↔ H2O(g) Slide4: Intermolecular ENERGY Interaction Type (kJ/mol) Ion – Ion 300 – 600 Ion – Dipole 10 – 100 Dipole – Dipole 1 – 10 H-Bonds 20 – 40 Ion – Induced Dipole 1 – 10 Dipole – Induced Dipole* 0.05 – 10 Induced dipole* – Induced dipole* 0.05 – 2 *Induced dipole forces also called London Dispersion ForcesSlide6: Section 9.2 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.Slide7: Atomic radii for cations of main group elements are smaller than the anions for these elements. What is the explanation for this general observation?Slide8: 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. Slide9: Coulomb’s Law: E = 2.31E -21 J۰nm (Q1Q2/d) Lattice Energy U = k (Q1Q2/d) Problem 22. Rank the following compounds in order of increasing melting point: BaCl2, CaCl2, MgCl2, and SrCl2. Note: melting points (or boiling points) for ionic compounds should increase with increasing intermolecular interactions (lattice energies).Slide10: Atomic radii for anions of main group elements are larger than the neutral species for these elements. What is the explanation for this general observation?Formally the lattice energy is the energy released when a mole of the ionic compound forms from the ions in the gas phase (Step 5).: Formally the lattice energy is the energy released when a mole of the ionic compound forms from the ions in the gas phase (Step 5).Formally the lattice energy is the energy released when a mole of the ionic compound forms from the free ions in the gas phase.: Formally the lattice energy is the energy released when a mole of the ionic compound forms from the free ions in the gas phase.Lattice Energy Tutorial : Learn to apply Coulomb's law to calculate the lattice energies of ionic solids. Includes practice exercises. »PC version Lattice Energy Tutorial Slide14: Problem 23 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 For further practice, work Problem 24 in the text.Slide15: Problem 24 Calculate the lattice energy of sodium oxide from the following data: Ionization energy of Na = 795 kJ/mol Electron Affinity of O for 2 e- = 638 kJ/mol Energy to vaporize Na = 109 kJ/mol O2 bond energy = 499 kJ/mol Energy change for the reaction: 2 Na(s) + ½ O2(g) → Na2O(s) is -416 kJ/mol For further understanding, look at Sample Ex. 9.3 in the text. Btw…the answer reported in the text should be negative!!Slide18: Section 9.3 Interactions of Polar Molecules In aqueous solution, ions are “hydrated” i.e. surrounded by water molecules due to ion-dipole interactions.Slide19: The figure below also shows intermolecular interaction between the water (solvent) molecules. These interaction are known as dipole-dipole interactions.Non-Polar molecules such as oxygen or nitrogen can interact by dispersive, or London forces (Induced dipoles). : Non-Polar molecules such as oxygen or nitrogen can interact by dispersive, or London forces (Induced dipoles). Contact between two non-polar molecule induces a short-lived dipole momentIntermolecular Forces Tutorial : This tutorial 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. »PC version Intermolecular Forces Tutorial Slide24: Physical Properties influenced by intermolecular forces: State (i.e. solid, liquid, or gas) under standard conditions. Boiling point; melting point. Miscibility Vapor pressure Slide26: 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”.Slide27: 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”.Slide28: Periodic trends in the boiling points of binary hydrogen compounds. Notice that H2O, HF, and NH3 are exceptions to the trends.Slide29: Why are water and 2-pentanone immiscible?Slide30: Which vitamins are water soluble and which are “fat” soluble?Slide31: O2 dissolves in water due to the interaction of an induced dipole moment of the O2 with the dipolar water molecule. This is an example of dipole-induced dipole interaction. Slide32: In a closed container, the lower Vapor Pressure of seawater leads to a transfer of distilled water to the beaker containing seawater.Slide33: Vapor Pressure Schematic view of evaporation and condensation in a closed system. (a) Initial conditions. (b) evaporation (why?). (c) Equilibrium conditions: partial pressure = vapor pressure.Slide34: Why does the Vapor pressure of a pure liquid increase with temperature? Then normal boiling point is defined as the temperature at which the vapor pressure of the liquid equals 1 atm.Slide35: Problem 35. Rank the following compounds in order of increasing vapor pressure at 298 K: CH3CH2OH CH3OCH3 CH3CH2CH3Slide36: Problem 35. Rank the following compounds in order of increasing vapor pressure at 298 K: CH3CH2OH CH3OCH3 CH3CH2CH3 BPs 79°C -25°C -42°CRaoult’s Law Tutorial : Explore the connection between the vapor pressure of a solution and its concentration as a gas above the solution. Includes practice exercises. »PC version Raoult’s Law Tutorial Slide38: Vapor Pressure Problem Suppose that 651 g of ethylene glycol, HOCH2CH2OH (MW= 62.07), is dissolved in 1.50 kg of water. What is the vapor pressure over the solution at 90°C? The vapor pressure of pure water is 525.8 mm Hg at 90°C. Slide39: Phase Diagram for WaterSlide40: The density of water reaches a maximum at 4°C, and decreases as the temperature is lowered. The phase diagram suggests that ice will melt when the pressure is increased.Phase Diagram for Carbon Dioxide: Phase Diagram for Carbon DioxidePhase Diagrams Tutorial : Use an interactive phase diagram and animated heating curve to explore how changes in temperature and pressure affect the physical state of a substance. »PC version Phase Diagrams Tutorial Hydrogen Bonding in Water Tutorial : This unit provides interactive 3D representations of water molecules, illustrates how the arrangement of the molecules in water and ice differ, and shows the molecular details of oil droplet formation in water. » Run Animation Hydrogen Bonding in Water Tutorial * This tutorial requires Netscape 4.08 and the MDL-Chime plug-in Slide44: Surface tension and adhesion properties of water.Slide45: Surface tension, viscosity and adhesion properties are common properties of liquids that are affected by intermolecular interactions. Viscosity is the resistance to flow of the liquid, and has units: kg۰m-1۰s-1 Why do certain oils and sugar solutions have large viscosities? Would you expect the viscosities of salt solutions to be larger or small than that of pure water? Capillary Action Tutorial : This tutorial demonstrates 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. »PC version Capillary Action Tutorial 09_16_n.jpg: 09_16_n.jpg09_19.jpg: 09_19.jpgCQ9-12.1a-Immersion of Ag Rod in Acid Solution: CQ9-12.1a-Immersion of Ag Rod in Acid Solution Given that the standard reduction potential of Ag+(aq) is 0.80 V, will a silver rod dissolve in an acidic solution of pH 0.0 under standard conditions? A) Yes B) No C) Depends CQ9-12.1b-Immersion of Ag Rod in Acid Solution: CQ9-12.1b-Immersion of Ag Rod in Acid Solution Consider the following arguments for each answer and vote again: All metals, including Ag(s), will dissolve in concentrated acid solution. The standard reduction potential of H+(aq) is less than that of Ag+(aq), so Ag(s) does not dissolve in an acidic solution. Whether the silver rod dissolves depends on whether a strong acid or a weak acid is used.W. W. Norton & CompanyIndependent and Employee-Owned: W. W. Norton & Company Independent and Employee-Owned This concludes the Norton Media Library slide set for chapter 9 Chemistry The Science in Context by Thomas Gilbert, Rein V. Kirss, & Geoffrey Davies You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.
chapter09 Davidino Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINTLite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 3215 Category: Education License: All Rights Reserved Like it (0) Dislike it (0) Added: January 08, 2008 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript ChemistryThe Science in ContextChapter 9Intermolecular Forces and Liquids : Chemistry The Science in Context Chapter 9 Intermolecular Forces and Liquids Slide2: 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. Slide3: Section 9.1 Why is NaCl more soluble in water than CaSO4? Substances can exist in three states: solid, liquid, and gas phases. In which of these phases are there the greatest number of intermolecular interactions? H2O(s) ↔ H2O(l) ↔ H2O(g) Slide4: Intermolecular ENERGY Interaction Type (kJ/mol) Ion – Ion 300 – 600 Ion – Dipole 10 – 100 Dipole – Dipole 1 – 10 H-Bonds 20 – 40 Ion – Induced Dipole 1 – 10 Dipole – Induced Dipole* 0.05 – 10 Induced dipole* – Induced dipole* 0.05 – 2 *Induced dipole forces also called London Dispersion ForcesSlide6: Section 9.2 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.Slide7: Atomic radii for cations of main group elements are smaller than the anions for these elements. What is the explanation for this general observation?Slide8: 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. Slide9: Coulomb’s Law: E = 2.31E -21 J۰nm (Q1Q2/d) Lattice Energy U = k (Q1Q2/d) Problem 22. Rank the following compounds in order of increasing melting point: BaCl2, CaCl2, MgCl2, and SrCl2. Note: melting points (or boiling points) for ionic compounds should increase with increasing intermolecular interactions (lattice energies).Slide10: Atomic radii for anions of main group elements are larger than the neutral species for these elements. What is the explanation for this general observation?Formally the lattice energy is the energy released when a mole of the ionic compound forms from the ions in the gas phase (Step 5).: Formally the lattice energy is the energy released when a mole of the ionic compound forms from the ions in the gas phase (Step 5).Formally the lattice energy is the energy released when a mole of the ionic compound forms from the free ions in the gas phase.: Formally the lattice energy is the energy released when a mole of the ionic compound forms from the free ions in the gas phase.Lattice Energy Tutorial : Learn to apply Coulomb's law to calculate the lattice energies of ionic solids. Includes practice exercises. »PC version Lattice Energy Tutorial Slide14: Problem 23 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 For further practice, work Problem 24 in the text.Slide15: Problem 24 Calculate the lattice energy of sodium oxide from the following data: Ionization energy of Na = 795 kJ/mol Electron Affinity of O for 2 e- = 638 kJ/mol Energy to vaporize Na = 109 kJ/mol O2 bond energy = 499 kJ/mol Energy change for the reaction: 2 Na(s) + ½ O2(g) → Na2O(s) is -416 kJ/mol For further understanding, look at Sample Ex. 9.3 in the text. Btw…the answer reported in the text should be negative!!Slide18: Section 9.3 Interactions of Polar Molecules In aqueous solution, ions are “hydrated” i.e. surrounded by water molecules due to ion-dipole interactions.Slide19: The figure below also shows intermolecular interaction between the water (solvent) molecules. These interaction are known as dipole-dipole interactions.Non-Polar molecules such as oxygen or nitrogen can interact by dispersive, or London forces (Induced dipoles). : Non-Polar molecules such as oxygen or nitrogen can interact by dispersive, or London forces (Induced dipoles). Contact between two non-polar molecule induces a short-lived dipole momentIntermolecular Forces Tutorial : This tutorial 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. »PC version Intermolecular Forces Tutorial Slide24: Physical Properties influenced by intermolecular forces: State (i.e. solid, liquid, or gas) under standard conditions. Boiling point; melting point. Miscibility Vapor pressure Slide26: 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”.Slide27: 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”.Slide28: Periodic trends in the boiling points of binary hydrogen compounds. Notice that H2O, HF, and NH3 are exceptions to the trends.Slide29: Why are water and 2-pentanone immiscible?Slide30: Which vitamins are water soluble and which are “fat” soluble?Slide31: O2 dissolves in water due to the interaction of an induced dipole moment of the O2 with the dipolar water molecule. This is an example of dipole-induced dipole interaction. Slide32: In a closed container, the lower Vapor Pressure of seawater leads to a transfer of distilled water to the beaker containing seawater.Slide33: Vapor Pressure Schematic view of evaporation and condensation in a closed system. (a) Initial conditions. (b) evaporation (why?). (c) Equilibrium conditions: partial pressure = vapor pressure.Slide34: Why does the Vapor pressure of a pure liquid increase with temperature? Then normal boiling point is defined as the temperature at which the vapor pressure of the liquid equals 1 atm.Slide35: Problem 35. Rank the following compounds in order of increasing vapor pressure at 298 K: CH3CH2OH CH3OCH3 CH3CH2CH3Slide36: Problem 35. Rank the following compounds in order of increasing vapor pressure at 298 K: CH3CH2OH CH3OCH3 CH3CH2CH3 BPs 79°C -25°C -42°CRaoult’s Law Tutorial : Explore the connection between the vapor pressure of a solution and its concentration as a gas above the solution. Includes practice exercises. »PC version Raoult’s Law Tutorial Slide38: Vapor Pressure Problem Suppose that 651 g of ethylene glycol, HOCH2CH2OH (MW= 62.07), is dissolved in 1.50 kg of water. What is the vapor pressure over the solution at 90°C? The vapor pressure of pure water is 525.8 mm Hg at 90°C. Slide39: Phase Diagram for WaterSlide40: The density of water reaches a maximum at 4°C, and decreases as the temperature is lowered. The phase diagram suggests that ice will melt when the pressure is increased.Phase Diagram for Carbon Dioxide: Phase Diagram for Carbon DioxidePhase Diagrams Tutorial : Use an interactive phase diagram and animated heating curve to explore how changes in temperature and pressure affect the physical state of a substance. »PC version Phase Diagrams Tutorial Hydrogen Bonding in Water Tutorial : This unit provides interactive 3D representations of water molecules, illustrates how the arrangement of the molecules in water and ice differ, and shows the molecular details of oil droplet formation in water. » Run Animation Hydrogen Bonding in Water Tutorial * This tutorial requires Netscape 4.08 and the MDL-Chime plug-in Slide44: Surface tension and adhesion properties of water.Slide45: Surface tension, viscosity and adhesion properties are common properties of liquids that are affected by intermolecular interactions. Viscosity is the resistance to flow of the liquid, and has units: kg۰m-1۰s-1 Why do certain oils and sugar solutions have large viscosities? Would you expect the viscosities of salt solutions to be larger or small than that of pure water? Capillary Action Tutorial : This tutorial demonstrates 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. »PC version Capillary Action Tutorial 09_16_n.jpg: 09_16_n.jpg09_19.jpg: 09_19.jpgCQ9-12.1a-Immersion of Ag Rod in Acid Solution: CQ9-12.1a-Immersion of Ag Rod in Acid Solution Given that the standard reduction potential of Ag+(aq) is 0.80 V, will a silver rod dissolve in an acidic solution of pH 0.0 under standard conditions? A) Yes B) No C) Depends CQ9-12.1b-Immersion of Ag Rod in Acid Solution: CQ9-12.1b-Immersion of Ag Rod in Acid Solution Consider the following arguments for each answer and vote again: All metals, including Ag(s), will dissolve in concentrated acid solution. The standard reduction potential of H+(aq) is less than that of Ag+(aq), so Ag(s) does not dissolve in an acidic solution. Whether the silver rod dissolves depends on whether a strong acid or a weak acid is used.W. W. Norton & CompanyIndependent and Employee-Owned: W. W. Norton & Company Independent and Employee-Owned This concludes the Norton Media Library slide set for chapter 9 Chemistry The Science in Context by Thomas Gilbert, Rein V. Kirss, & Geoffrey Davies