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Liquids and Solids

Liquids and Solids. Ch 10.2,10.3 & 10.4 Pg. 353 # 4, 5, 7, 8, 10-14,17, 20-22, 27, 28, 33. Liquids exist in the smallest temperature range, so liquids are the least common state of matter. Kinetic Theory Description of the Liquid State

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Liquids and Solids

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  1. Liquids and Solids Ch 10.2,10.3 & 10.4 Pg. 353 # 4, 5, 7, 8, 10-14,17, 20-22, 27, 28, 33

  2. Liquids exist in the smallest temperature range, so liquids are the least common state of matter . . . • Kinetic Theory Description of the Liquid State • According to the kinetic theory, motion of liquid particles can be described as . . .

  3. Definite Volume Fluidity Relative High Density Incompressible Dissolving Ability Ability to Diffuse Surface Tension Tendency to Evaporate and Boil Tendency to Solidify Properties of Liquids and the Particles Model – define each property Properties of Fluids

  4. 13.1 Questions • Why are liquids more dense than gases? • Molecules are closer together so more molecules in a given area • Why are liquids harder to compress than gases? • Same as above – molecules are closer • Why do liquids diffuse slower than gases? • Particles are not moving as fast as gases • Can a liquid boil without increasing the temperature? How? • Yes – lower the atmospheric pressure

  5. 10.3 Solids • “Solid as a rock, “ is the description of solid – something that is hard, unyielding, with a definite shape and volume. Many things other than rocks are solids. In fact, solids are more common than liquids. This diagram shows the particles of a gas, liquid and solid.

  6. Kinetic-Theory Description of the Solid StateAccording to the kinetic theory, the motion of solid particles can be described as…. • Lower kinetic energy, less motion, more packed particles, and higher intermolecular forces (IMF) • Properties of Solids and the Particle Model – define each property: Properties of Solids • Definite shape and volume • Non-fluid • Definite melting point • High Density • Incompressible • Slow Diffusion

  7. Crystalline Solids • Classification of crystals by arrangement and shape • Crystal Lattice (define) - The total 3-D array of points that describe the arrangement of the particles – a collection of unit cells. • The smallest portion of the crystal lattice that reveals the 3-D pattern of the entire lattices is the unit cell.

  8. Binding Forces in Crystals Body-centered (ex. Li, K, Cr) Simple

  9. Types of Crystals Hexogonal (like oranges in a grocery store); (ex. Zn) Face-centered (ex. Cu, Ag, Au)

  10. Binding forces in crystals

  11. Amorphous Solids • Rubber, glass, plastics and synthetic fibers are called amorphous solids.

  12. “Amorphous,” comes from the Greek for “without a shape.” • Unlike crystals, amorphous solids do not have a regular, natural shape, but instead take on whatever shape imposed on them. • Particle arrangement is not uniform; they are arranged randomly, like particles of a liquid. • Examples of amorphous solids – glass used in fiberoptics (optical fibers transmit telephone conversations by means of light waves.

  13. Amorphous solids are prepared by rapid cooling of thin film materials. • Molecular examples Crystalline vs. Amorphous

  14. 10.4 Changes of State

  15. Equilibrium • What does equilibrium mean? • It is a dynamic condition in which two opposing changes occur at equal rates in a closed system. • What is a closed system? • H2O in an open beaker • H2O in a closed beaker

  16. When a liquid changes to a vapor, as in evaporation, it absorbs heat energy and can be shown as: • Open system evaporation – liquid + heat vapor • Closed system evaporation – liquid + heat  vapor • When a vapor condenses, as in condensation, it gives off heat energy and can be shown as: • And condensation – vapor  liquid + heat • The liquid vapor equilibrium can be rewritten as: • liquid + heat↔vapor • “The double yields sign represents a reaction at equilibrium”

  17. Le Chatelier’s Principle • What is it? LeChatelier • When a system at equilibrium is disturbed by the application of stress, the system reacts to minimize the stress. • Is temperature an example of stress? • Yes. • What happens when you increase the temperature of a system? Equ. shift from heat • ↓ liquid + increased heat ---> ↑ vapor

  18. Le Chatelier’s Principle • What happens when you decrease the temperature of a system? • ↓ vapor ---> ↑ liquid + decreased heat • What factor is controlling the decrease and increase of vapor and liquid? • the temperature (heat)

  19. Equilibrium Vapor Pressure of a Liquid • What is it? • At equilibrium, the molecules of a vapor exert a specific pressure on its corresponding liquid.

  20. When equilibrium vapor pressure of water is graphed, (draw figure 14 below):

  21. The strength of attractive forces is independent of temperature. Higher temperatures with resultant higher kinetic energies make these forces less effective. • Liquid water can exist in equilibrium with water vapor only up to a temperature of 374.1ºC. Later you will learn that neither liquid water nor water vapor can exist at temperatures above 374.1ºC.

  22. What is equilibrium called when liquid molecules enter into the gaseous state? • Vaporization • Where does this occur? • On the surface of the liquid = evaporation, throughout liquid = boiling • Equilibrium vapor pressure depends on: • a) temperature and pressure • b) boiling point of a liquid (the type of liquid)

  23. If a liquid has high intermolecular forces, then what happens to that liquid’s vapor pressure? Why? • vapor pressure ↓ high IMFs = increase hold on the molecules

  24. Boiling. Freezing. Melting • What is boiling? • The conversion of a liquid to a vapor, within the liquid as well as its surface when the equilibrium vapor pressure of the liquid is equal to the atmospheric pressure. • What is the boiling point? • The temperature at which the equilibrium vapor pressure of the liquid is equal to the atmospheric pressure (760 torr). • Boiling happens throughout the liquid…evaporation happens on the surface.

  25. What is the molar heat of vaporization? • The amount of heat energy required to vaporize one mole of liquid at its boiling point. • How does a pressure cooker work? • It elevates pressure to raise boiling point and shorten cooking time.

  26. Freezing and melting • What is the freezing? • The physical change of a liquid to a solid. • What is melting? • The physical change of a solid to liquid. • What is the molar heat of fusion? • The amount of heat energy required to melt one mole of solid at its melting point.

  27. solid + heat  Liquid liquid solid + heat re-write the equation: solid + heat ↔ liquid heat of fusion

  28. Are the freezing points and melting points the same temperature? • Yes • at 0°C H2O with 6kJ is a liquid • at 0°C H2O without 6kJ is a solid

  29. Chapter 10 Calculations – not in book • Molar heat of Vaporization • The amount of heat energy required to vaporize one mole of liquid at its boiling point. • Joules are the standard unit to measure heat energy. • Molar heat of vaporization for water is 40.79 kJ/mole.

  30. 2.2 – Heat and Temperature – there is a difference • Heat transfers between objects – flows from hot to cold - Law of Conservation of Energy • Ex1:ice cube in a thermos of hot water - ice melts, water cools - same amount of heat • SI unit of heat - Joule (J) calorie is also used frequently • Calorie - the amount of energy required to raise the temperature of 1 g of water by 1 oC • (Calories – capital letter – really means kilocalories – used in food energy measurement) 1.000 calorie = 4.184 Joules

  31. Specific Heat Problems • For water, Cp = 1.000 cal/goC or 4.184 J/goC for water  • Ex1:How many calories does it take to heat 20. g of water from 10.0 to 40.0oC? Also how many J? • Ex2: How much heat is required to heat 75 g of Iron (Cp = 0.444 J/gCo) from 15.5 to 57.0oC?

  32. Specific Heat Problems • Ex3: What is the specific heat of an object if 250 calories will heat 55 g of it from 25 to 100.0 oC? • Ex4: - If a 100.0 g sample of silver (Cp = .237 J/g oC) at 80.0 Co loses 50. calories, what will its final temperature be?

  33. Not In Book • NIB: It also takes energy to melt or boil any substance. The amount of energy required to melt or boil a substance can be expressed by the following equations:  • ΔH = nΔHfusion ΔH = change in energy (J) n = number of moles • ΔH = nΔHvaporizationΔHfusion = the molar heat of fusion (kJ/mol) ΔHvaporization = the molar heat of vaporization (kJ/mol) • ΔHfusion and ΔHfusion are constants and correspond to the amount of energy it takes to freeze (fuse) or boil (vaporize) one mole of a substance. • When doing heat calculations that involve both a change of state and a change in temperature, make sure the answers for both calculations are written in the same units before adding them together!

  34. Ex1: How much heat energy would be required to vaporize 5.00 moles of H2O • q = ΔHvap·(mol) = 40.79 kJ/mol · 5.00 mol = 204 kJ or 204,000 J • Ex2: to vaporize 45.0g of H2O • q = ΔHvap·(mol) 45.0g ·1mol = 40.79 kJ/mol (2.50 mol) 1 18.0g = 102 kJ or 102,000 J

  35. when....a liquid evaporates, it absorbs energy. Energy is used to overcome attractive forces.The energy doesn’t increase the average energy of the particles, so the temperature doesn’t change. • when...a liquid evaporates, it takes energy from its surroundings that’s why alcohol feels cool to the skin. • it’s also why we get cold when getting out of the shower

  36. Heat of vaporization - Hvap - energy needed to vaporize a unit of substance (mass or moles) • Formula: q = (H vap ) x ( unit ) unit = gram or mole • Ex3 - How much heat does it take to vaporize 50.0 g of water at 100.0 °C 50.0g · 1mol = 2.78 mol 1 18.0g • q = (H vap ) x ( unit ) = 40.79 (2.78) = 113 kJ

  37. Molar Heat of Fusion • The amount of heat energy required to melt one mole of a solid at its melting point. • The molar heat of fusion of water is 6.008 kJ/mole. • Ex1: How much energy would be required to melt 12.75 moles of ice? • q = ΔHfus·(mol) = 6.008 kJ/mol ·(12.75 mol) = 76.60 kJ

  38. Ex2: to melt 6.48 x 1020 kg of ice? • 6.48x1020kg· 1000 g = 6.48x1023 g 1 1kg 6.48x1023 g · 1mol = 3.6x1022 mol 1 18.0g 6.008kJ/mol(3.6x1022 mol) = 2.16x1023 kJ

  39. Heat of Fusion - Hfus = heat of fusion - heat required to change a unit of substance from solid to liquid • same formula: q = (Hfus) x (unit) unit = g or mole • Ex3: - How much ice can be melted by 2.9 x 104 J? • 2.9x104J· 1kJ = 29kJ 1 1000J q = (Hfus) x (mol) 29 kJ = 6.008 kJ/mol x (mol) = 4.8 mol ice

  40. Temperature and Phase Changes

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