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2 Energy and Matter Unit 2

3 Chemical Change: Any change involving a rearrangement of atoms.

4 During a “chemical reaction” new materials are formed by a change in the way atoms are bonded together. During a “chemical reaction” new materials are formed by a change in the way atoms are bonded together.

5 Physical and Chemical Properties Examples of Physical Properties Boiling point Color SlipperinessElectrical conductivity Melting point TasteOdorDissolves in water Shininess (luster) SoftnessDuctilityViscosity (resistance to flow) Volatility HardnessMalleabilityDensity (mass / volume ratio) Examples of Chemical Properties Burns in air Reacts with certain acidsDecomposes when heated Explodes Reacts with certain metalsReacts with certain nonmetals Tarnishes Reacts with waterIs toxic Ralph A. Burns, Fundamentals of Chemistry 1999, page 23 Chemical properties can ONLY be observed during a chemical reaction!

6 Three Possible Types of Bonds + - ++ -- Covalent e.g. H 2 Polar Covalent e.g. HCl Ionic e.g. NaCl

7 Shattering an Ionic Crystal; Bending a Metal Bailar, Jr, Moeller, Kleinberg, Guss, Castellion, Metz, Chemistry, 1984, page An ionic crystal A metal No electrostatic forces of repulsion – metal is deformed (malleable) Electrostatic forces of repulsion Force broken crystal

8 Chemical Bonds Increasing ionic character Nonpolar covalent Electrons are shared equally Cl Polar covalent Electrons are shared unequally Cl H Ionic bonding Electrons are transferred Cl 1- Na 1+ Ralph A. Burns, Fundamentals of Chemistry 1999, page 229 between two identical nonmetal atoms are nonpolar covalent. between two different nonmetal atoms are polar covalent. between nonmetals and reactive metals are primarily ionic.

9 Covalent vs. Ionic Covalent Transfer electrons (ions formed) + / - Between Metal and Nonmetal Strong Bonds (high melting point) Share electrons (polar vs. nonpolar) Between Two Nonmetals Weak Bonds (low melting point) Alike Different Electrons are involved Chemical Bonds Ionic Different Topic

10 Temperature Scales Fahrenheit 212 o F 180 o F 32 o F Celcius 100 o C 0 o C Kelvin 373 K 100 K 273 K Boiling point of water Freezing point of water 1 kelvin degree = 1 degree Celcius Notice that 1 kelvin degree = 1 degree Celcius

11 Heat versus Temperature Kinetic energy Fractions of particles lower temperature higher temperature TOTAL Kinetic ENERGY = Heat

12 Molecular Velocities speed Fractions of particles many different molecular speeds molecules sorted by speed the Maxwell speed distribution

13 Temperature vs. Heat Measured with a Calorimeter Total Kinetic Energy Joules (calories) Measured with a Thermometer Average Kinetic Energy o Celcius (or Kelvin) Alike Different A Property of Matter Have Kinetic Energy Heat Different Topic Temperature thermometer calorimeter

14 Conservation of Matter Reactants yield Products

15 Density Density is an INTENSIVE INTENSIVE property of matter. - does NOT depend on quantity of matter. - color, melting point, boiling point, odor, density Contrast with EXTENSIVE - depends on quantity of matter. - mass, volume, heat content (calories) Styrofoam Brick

16 Properties of Matter Pyrex Pyrex Extensive Properties Intensive Properties volume: mass: density: temperature: 100 mL g g/mL 20 o C 15 mL g g/mL 20 o C

17 Volume and Density Relationship Between Volume and Density for Identical Masses of Common Substances Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm 3 ) (g/cm 3 ) Lithium Water Aluminum Lead

18 Density D M V ensity ass olume D = M VM V M = D x V V = M DM D

19 Two ways of viewing density Dorin, Demmin, Gabel, Chemistry The Study of Matter, 3 rd Edition, 1990, page 71 Equal volumes… …but unequal masses The more massive object (the gold cube) has the greater density. aluminum gold (A) Equal masses… …but unequal volumes. (B) gold aluminum The object with the larger volume (aluminum cube) has the smaller density.

20 Specific Gravity Jaffe, New World of Chemistry, 1955, page water 1.0 ice cork aluminum 2.7

21 Archimedes Principle V final = 98.5 cm 3 - V initial = 44.5 cm 3 V fishing sinker = 54.0 cm 3 Before immersion Water 44.5 cm 3 After immersion Fishing sinker 98.5 cm 3 Thread

22 Dissolving of Salt in Water NaCl(s) + H 2 O  Na + (aq) + Cl - (aq)

23 Liquids The two key properties we need to describe are EVAPORATIONCONDENSATION EVAPORATION and its opposite CONDENSATION add energy and break intermolecular bonds EVAPORATION release energy and form intermolecular bonds CONDENSATION

24 States of Matter Solid Liquid Gas Holds Shape Fixed Volume Shape of Container Free Surface Fixed Volume Shape of Container Volume of Container heat

25 Some Properties of Solids, Liquids, and Gases Property Solid Liquid Gas Shape Has definite shapeTakes the shape of Takes the shape the container of its container Volume Has a definite volumeHas a definite volume Fills the volume of the container Arrangement of Fixed, very closeRandom, close Random, far apart Particles Interactions between Very strongStrong Essentially none particles

26 To evaporate, molecules must have sufficient energy to break IM forces. Molecules at the surface break away and become gas. Only those with enough KE escape. endothermicBreaking IM forces requires energy. The process of evaporation is endothermic. Evaporation is a cooling process. It requires heat. Evaporation

27 Change from gas to liquid Achieves a dynamic equilibrium with vaporization in a closed system. What is a closed system? A closed system means matter can’t go in or out. (put a cork in it) What the heck is a “dynamic equilibrium?” Condensation

28 When first sealed, the molecules gradually escape the surface of the liquid. As the molecules build up above the liquid - some condense back to a liquid. The rate at which the molecules evaporate and condense are equal. Dynamic Equilibrium

29 As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense. Equilibrium is reached when: Rate of Vaporization = Rate of Condensation Molecules are constantly changing phase “dynamic” The total amount of liquid and vapor remains constant “equilibrium” Dynamic Equilibrium

30 Vaporization is an endothermic process - it requires heat. Energy is required to overcome intermolecular forces Responsible for cool earth Why we sweat Vaporization

31 Energy Changes Accompanying Phase Changes Solid Liquid Gas Melting Freezing Deposition CondensationVaporization Sublimation Energy of system Brown, LeMay, Bursten, Chemistry  2000, page 405

32 solid liquid gas Heat added Temperature ( o C) A B C D E Heating Curve for Water LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World, 1996, page 487

33 solid liquid gas vaporization condensation melting freezing Heat added Temperature ( o C) A B C D E Heating Curve for Water LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World, 1996, page 487

34 Latent Heat Take 1 kg of water from –10 o C up to 150 o C we can plot temperature rise against absorbed heat water steam (water vapor) -10 C 0 C 100 C ice L f = 80 cal/gL v = 540 cal/g L f is the latent heat of fusion L v is the latent heat of vaporization Q heat absorbed

35 MATTER Can it be physically separated? Homogeneous Mixture (solution) Heterogeneous MixtureCompoundElement MIXTUREPURE SUBSTANCE yesno Can it be chemically decomposed? noyes Is the composition uniform? noyes ColloidsSuspensions Courtesy Christy Johannesson

36 Elements only one kind of atom; atoms are bonded if the element is diatomic or polyatomic Compounds two or more kinds of atoms that are bonded substance with definite makeup and properties Mixtures two or more substances that are physically mixed two or more kinds of and Both elements and compounds have a definite makeup and definite properties. Packard, Jacobs, Marshall, Chemistry Pearson AGS Globe, page (Figure 2.4.1)

37 Matter Flowchart Examples: –graphite –pepper –sugar (sucrose) –paint –soda Courtesy Christy Johannesson element hetero. mixture compound solution homo. mixture hetero. mixture

38 Pure Substances Element –composed of identical atoms –examples: copper wire, aluminum foil Courtesy Christy Johannesson

39 Pure Substances Compound –composed of 2 or more elements in a fixed ratio –properties differ from those of individual elements –EX: table salt (NaCl) Courtesy Christy Johannesson

40 Pure Substances Law of Definite Composition –A given compound always contains the same, fixed ratio of elements. Law of Multiple Proportions –Elements can combine in different ratios to form different compounds. Courtesy Christy Johannesson

41 Mixtures Variable combination of two or more pure substances. HeterogeneousHomogeneous Courtesy Christy Johannesson

42 Mixtures Solution –homogeneous –very small particles –no Tyndall effect Tyndall Effect –particles don’t settle –EX: rubbing alcohol Courtesy Christy Johannesson

43 Mixtures Colloid –heterogeneous –medium-sized particles –Tyndall effect –particles don’t settle –EX: milk Courtesy Christy Johannesson

44 Mixtures Suspension –heterogeneous –large particles –Tyndall effect –particles settle –EX: fresh-squeezed lemonade Courtesy Christy Johannesson

45 Classification of Matter Materials Homogeneous Heterogeneous mixture Homogeneous mixture Substance ElementCompoundSolutionMixture Specific / General Order / Disorder Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 43

46 Classification of Matter MATTER (gas. Liquid, solid, plasma) PURE SUBSTANCES MIXTURES HETEROGENEOUS MIXTURE HOMOGENEOUS MIXTURES ELEMENTSCOMPOUNDS Separated by physical means into Separated by chemical means into Kotz & Treichel, Chemistry & Chemical Reactivity, 3 rd Edition, 1996, page 31

47 Classification of Matter uniform properties? fixed composition? chemically decomposable? no yes hetero- geneous mixture solution element compound

48 Elements, Compounds, and Mixtures (a) an element (hydrogen) (b) a compound (water) (c) a mixture (hydrogen and oxygen) (d) a mixture (hydrogen and oxygen) Dorin, Demmin, Gabel, Chemistry The Study of Matter, 3 rd Edition, 1990, page 68 hydrogen atoms hydrogen atoms oxygen atoms

49 Mixture vs. Compound Mixture Fixed Composition Bonds between components Can ONLY be separated by chemical means Variable Composition No bonds between components Can be separated by physical means Alike Different Contain two or more elements Can be separated into elements Involve substances Compound Different Topic

50 Compounds vs. Mixtures Compounds have properties that are uniquely different from the elements from which they are made. –A formula can always be written for a compound –e.g. NaCl  Na + Cl 2 Mixtures retain their individual properties. –e.g. Salt water is salty and wet

51 Diatomic Elements, 1 and 7 H2H2 N2N2 O2O2 F2F2 Cl 2 Br 2 I2I2

52 Matter Substance Definite composition (homogeneous) Substance Definite composition (homogeneous) Element (Examples: iron, sulfur, carbon, hydrogen, oxygen, silver) Element (Examples: iron, sulfur, carbon, hydrogen, oxygen, silver) Mixture of Substances Variable composition Mixture of Substances Variable composition Compound (Examples: water. iron (II) sulfide, methane, Aluminum silicate) Compound (Examples: water. iron (II) sulfide, methane, Aluminum silicate) Homogeneous mixture Uniform throughout, also called a solution (Examples: air, tap water, gold alloy) Homogeneous mixture Uniform throughout, also called a solution (Examples: air, tap water, gold alloy) Heterogeneous mixture Nonuniform distinct phases (Examples: soup, concrete, granite) Heterogeneous mixture Nonuniform distinct phases (Examples: soup, concrete, granite) Chemically separable Physically separable

53 The Organization of Matter MATTER PURE SUBSTANCES HETEROGENEOUS MIXTURE HOMOGENEOUS MIXTURES ELEMENTSCOMPOUNDS Physical methods Chemical methods Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 41

54 Phosphorous (P 4 ) TWO ALLOTROPIC FORMS White phosphorous spontaneously ignites Red phosphorous used for matches Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 457

55 Allotropes of Carbon GraphiteBuckminsterfullereneDiamond Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 27

56 Gold 24 karat gold 18 karat gold 14 karat gold Gold Copper Silver 18 / 24 atoms Au 24 / 24 atoms Au 14 / 24 atoms Au

57 Solid Brass An alloy is a mixture of metals. Brass = Copper + Zinc Solid brass homogeneous mixture a substitutional alloy Copper Zinc

58 Brass Plated Brass = Copper + Zinc Brass plated heterogeneous mixture Only brass on outside Copper Zinc

59 Galvanized Nails and Screws Zinc coating prevents rust –Use deck screws for any outdoor project Iron will rust if untreated –Weaken and break

60 Methods of Separating Mixtures Magnet Filter Decant Evaporation Centrifuge Chromatography Distillation

61 Chromatography Tie-dye t-shirt Black pen ink DNA testing –Tomb of Unknown Soldiers –Crime scene –Paternity testing

62 Paper Chromatography of Water-Soluble Dyes orange red yellow Initial spots of dyes Direction of Water (mobile phase) movement Filter paper (stationary phase) Orange mixture of red and yellow Suggested red dye is not homogeneous

63 Separation by Chromatography sample mixture a chromatographic column stationary phase selectively absorbs components mobile phase sweeps sample down column detector

64 Ion chromatogram of orange juice time (minutes) detector response Na + K+K+ Mg 2+ Fe 3+ Ca 2+

65 A Distillation Apparatus liquid with a solid dissolved in it thermometer condenser tube distilling flask pure liquid receiving flask hose connected to cold water faucet Dorin, Demmin, Gabel, Chemistry The Study of Matter, 3 rd Edition, 1990, page 282

66 Centrifugation Spin sample very rapidly: denser materials go to bottom (outside) Separate blood into serum and plasma –Serum (clear) –Plasma (contains red blood cells ‘RBCs’) Check for anemia (lack of iron) Blood RBC’s Serum A B C AFTER Before

67 The decomposition of two water molecules. 2 H 2 O  O H2H2 Electric current Water molecules Diatomic oxygen molecule hydrogen molecules +

68 Electrolysis *Must add acid catalyst to conduct electricity *H 1+ water oxygen hydrogen “electro” = electricity “lysis” = to split Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 32 Water Hydrogen gas forms Oxygen gas forms ElectrodeSource of direct current H 2 O(l) O 2 (g) + 2 H 2 (g)

69 Reviewing Concepts Reviewing Concepts Physical Properties List seven examples of physical properties. Describe three uses of physical properties. Name two processes that are used to separate mixtures. When you describe a liquid as thick, are you saying that it has a high or low viscosity?

70 Reviewing Concepts Reviewing Concepts Physical Properties Explain why sharpening a pencil is an example of a physical change. What allows a mixture to be separated by distillation?

71 Reviewing Concepts Reviewing Concepts Chemical Properties Under what conditions can chemical properties be observed? List three common types of evidence for a chemical change. How do chemical changes differ from physical changes?

72 Reviewing Concepts Reviewing Concepts Chemical Properties Explain why the rusting of an iron bar decreases the strength of the bar. A pat of butter melts and then burns in a hot frying pan. Which of these changes is physical and which is chemical?

73 ELEMENT hydrogen molecule, H 2 ELEMENT oxygen molecule, O 2 MIXTURE a mixture of hydrogen and oxygen molecules CHEMICAL REACTION if molecules collide with enough force to break them into atoms, a can take place COMPOUND water, H 2 O

74 2 H 2 O2O2 O2O2 2 H 2 O E E Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

75 Specific Heats of Some Substances Specific Heat Substance (cal/ g o C)(J/g o C) Water Alcohol Wood Aluminum Sand Iron Copper Silver Gold

76 Copyright © 2007 Pearson Benjamin Cummings. All rights reserved. (a) Radiant energy(b) Thermal energy (c) Chemical energy(d) Nuclear energy(e) Electrical energy

77 The energy something possesses due to its motion, depending on mass and velocity. Potential energy Energy in Energy out kinetic energy

78 Energy Kinetic Energy – energy of motion KE = ½ m v 2 Potential Energy – stored energy Batteries (chemical potential energy) Spring in a watch (mechanical potential energy) Water trapped above a dam (gravitational potential energy) massvelocity (speed) B A C

79 School Bus or Bullet? Which has more kinetic energy; a slow moving school bus or a fast moving bullet? Recall: KE = ½ m v 2 KE = ½ m v 2 BUSBULLET KE (bus) = ½ (10,000 lbs) (0.5 mph) 2 KE (bullet) = ½ (0.002 lbs) (240 mph) 2 Either may have more KE, it depends on the mass of the bus and the velocity of the bullet. Which is a more important factor: mass or velocity? Why?(Velocity) 2

80 Kinetic Energy and Reaction Rate Kinetic energy Fractions of particles lower temperature higher temperature minimum energy for reaction

81 Hot vs. Cold Tea Kinetic energy Many molecules have an intermediate kinetic energy Few molecules have a very high kinetic energy Low temperature (iced tea) High temperature (hot tea) Percent of molecules

82 Decomposition of Nitrogen Triiodide 2 NI 3 (s) N 2 (g) + 3 I 2 (g) NI 3 I2I2 N2N2

83 Exothermic Reaction Reactants  Products + Energy 10 energy = 8 energy + 2 energy Reactants Products -H-H Energy Energy of reactants Energy of products Reaction Progress

84 Endothermic Reaction Energy + Reactants  Products +  H Endothermic Reaction progress Energy Reactants Products Activation Energy

85 Effect of Catalyst on Reaction Rate reactants products Energy activation energy for catalyzed reaction Reaction Progress No catalyst Catalyst lowers the activation energy for the reaction. What is a catalyst? What does it do during a chemical reaction?

86 Burning of a Match Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 293 Energy released to the surrounding as heat SurroundingsSystem (Reactants)  (PE) Potential energy (Products)

87 Conservation of Energy in a Chemical Reaction Surroundings System Surroundings System Energy Before reaction After reaction In this example, the energy of the reactants and products increases, while the energy of the surroundings decreases. In every case, however, the total energy does not change. Myers, Oldham, Tocci, Chemistry, 2004, page 41 Endothermic Reaction Reactant + Energy Product

88 Direction of Heat Flow Surroundings ENDOthermic q sys > 0 EXOthermic q sys < 0 System Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 207 System H 2 O(s) + heat  H 2 O(l)melting H 2 O(l)  H 2 O(s) + heat freezing

89 Caloric Values Food joules/grams calories/gram Calories/gram Protein Fat Carbohydrates Smoot, Smith, Price, Chemistry A Modern Course, 1990, page calories = 1 Calorie "science" "food" 1calories = joules

90 Units of energy Most common units of energy 1. S  unit of energy is the joule (J), defined as 1 (kilogrammeter 2 )/second 2, energy is also expressed in kilojoules (1 kJ = 10 3 J). 2. Non-S  unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed to raise the temperature of 1 g of water by 1°C. One cal = J or 1J = cal. Units of energy are the same, regardless of the form of energy

91 Typical apparatus used in this activity include a boiler (such as large glass beaker), a heat source (Bunsen burner or hot plate), a stand or tripod for the boiler, a calorimeter, thermometers, samples (typically samples of copper, aluminum, zinc, tin, or lead), tongs (or forceps or string) to handle samples, and a balance. Experimental Determination of Specific Heat of a Metal

92 A Bomb Calorimeter

93 Heating Curves Temperature ( o C) Time Melting - PE  Solid - KE  Liquid - KE  Boiling - PE  Gas - KE 

94 Calculating Energy Changes - Heating Curve for Water Temperature ( o C) Time  H = mol x C fus  H = mol x C vap  H = mass x  T x C p, liquid  H = mass x  T x C p, gas  H = mass x  T x C p, solid C p, gas = 1.87 J/g o C C p, liquid = J/g o C C p, solid = J/g o C C f, water = 333 J/g C v, water = 2256 J/g

95 Heat Transfer Al m = 20 g T = 40 o C SYSTEM Surroundings m = 20 g T = 20 o C 20 g (40 o C) 20 g (20 o C)30 o C Block “A” Block “B” Final Temperature Assume NO heat energy is “lost” to the surroundings from the system. What will be the final temperature of the system ? a) 60 o C b) 30 o C c) 20 o C d) ?

96 Heat Transfer Al m = 20 g T = 40 o C SYSTEM Surroundings m = 10 g T = 20 o C 20 g (40 o C) 20 g (20 o C)30.0 o C Block “A” Block “B” Final Temperature Assume NO heat energy is “lost” to the surroundings from the system. 20 g (40 o C) 10 g (20 o C)33.3 o C What will be the final temperature of the system ? a) 60 o C b) 30 o C c) 20 o C d) ? ?

97 Heat Transfer Al m = 20 g T = 20 o C SYSTEM Surroundings m = 10 g T = 40 o C 20 g (40 o C) 20 g (20 o C)30.0 o C Block “A” Block “B” Final Temperature Assume NO heat energy is “lost” to the surroundings from the system. 20 g (40 o C) 10 g (20 o C)33.3 o C 20 g (20 o C) 10 g (40 o C)26.7 o C

98 Heat Transfer m = 75 g T = 25 o C SYSTEM Surroundings m = 30 g T = 100 o C 20 g (40 o C) 20 g (20 o C)30.0 o C Block “A” Block “B” Final Temperature 20 g (40 o C) 10 g (20 o C)33.3 o C 20 g (20 o C) 10 g (40 o C)26.7 o C Ag H2OH2O Real Final Temperature = 26.6 o C Why? We’ve been assuming ALL materials transfer heat equally well.

99 Specific Heat Water and silver do not transfer heat equally well. Water has a specific heat C p = J/g o C Silver has a specific heat C p = J/g o C What does that mean? It requires Joules of energy to heat 1 gram of water 1 o C and only Joules of energy to heat 1 gram of silver 1 o C. Law of Conservation of Energy… In our situation (silver is “hot” and water is “cold”)… this means water heats up slowly and requires a lot of energy whereas silver will cool off quickly and not release much energy. Lets look at the math!

100 “loses” heat Calorimetry m = 75 g T = 25 o C SYSTEM Surroundings m = 30 g T = 100 o C Ag H2OH2O T final = 26.6 o C

101 Calorimetry m = 75 g T = 25 o C SYSTEM Surroundings m = 30 g T = 100 o C Ag H2OH2O

102 1 Calorie = 1000 calories “food” = “science” Candy bar 300 Calories = 300,000 calories English Metric = _______ Joules 1 calorie - amount of heat needed to raise 1 gram of water 1 o C 1 calorie = Joules

103 C p (ice) = J/g o C It takes Joules to raise 1 gram ice 1 o C. X Joules to raise 10 gram ice 1 o C. (10 g)(2.077 J/g o C) = Joules X Joules to raise 10 gram ice 10 o C. (10 o C)(10 g)(2.077 J/g o C) = Joules Heat = (specific heat) (mass) (change in temperature) q = Cp. m.  T Temperature ( o C) Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid

104 Heat = (specific heat) (mass) (change in temperature) q = Cp. m.  T GivenT i = -30 o C T f = -20 o C q = Joules Temperature ( o C) Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid

105 240 g of water (initially at 20 o C) are mixed with an unknown mass of iron (initially at 500 o C). When thermal equilibrium is reached, the system has a temperature of 42 o C. Find the mass of the iron. Calorimetry Problems 2 question #5 Fe T = 500 o C mass = ? grams T = 20 o C mass = 240 g LOSE heat = GAIN heat - - [(C p, Fe ) (mass) (  T)] = (C p, H 2 O ) (mass) (  T) - [( J/g o C) (X g) (42 o C o C)] = (4.184 J/g o C) (240 g) (42 o C - 20 o C) Drop Units: - [(0.4495) (X) (-458)] = (4.184) (240 g) (22) X = X = g Fe

106 A 97 g sample of gold at 785 o C is dropped into 323 g of water, which has an initial temperature of 15 o C. If gold has a specific heat of J/g o C, what is the final temperature of the mixture? Assume that the gold experiences no change in state of matter. Calorimetry Problems 2 question #8 Au T = 785 o C mass = 97 g T = 15 o C mass = 323 g LOSE heat = GAIN heat - - [(C p, Au ) (mass) (  T)] = (C p, H 2 O ) (mass) (  T) - [(0.129 J/g o C) (97 g) (T f o C)] = (4.184 J/g o C) (323 g) (T f - 15 o C) Drop Units: - [(12.5) (T f o C)] = (1.35x 10 3 ) (T f - 15 o C) T f x 10 3 = 1.35 x 10 3 T f x x 10 4 = 1.36 x 10 3 T f T f = 22.1 o C

107 If 59 g of water at 13 o C are mixed with 87 g of water at 72 o C, find the final temperature of the system. Calorimetry Problems 2 question #9 T = 13 o C mass = 59 g LOSE heat = GAIN heat - - [(C p, H 2 O ) (mass) (  T)] = (C p, H 2 O ) (mass) (  T) - [(4.184 J/g o C) (59 g) (T f - 13 o C)] = (4.184 J/g o C) (87 g) (T f - 72 o C) Drop Units: - [(246.8) (T f - 13 o C)] = (364.0) (T f - 72 o C) T f = 364 T f = T f T f = 48.2 o C T = 72 o C mass = 87 g

108 A 38 g sample of ice at -11 o C is placed into 214 g of water at 56 o C. Find the system's final temperature. Calorimetry Problems 2 question #10 ice T = -11 o C mass = 38 g T = 56 o C mass = 214 g LOSE heat = GAIN heat - - [(C p, H 2 O ) (mass) (  T)] = (C p, H 2 O ) (mass) (  T) + (C f ) (mass) + (C p, H 2 O ) (mass) (  T) - [(4.184 J/g o C)(214 g)(T f - 56 o C)] = (2.077 J/g o C)(38 g)(11 o C) + (333 J/g)(38 g) + (4.184 J/g o C)(38 g)(T f - 0 o C) - [(895) (T f - 56 o C)] = (159) (T f ) T f = T f T f = T f T f = 34.7 o C = 1054 T f Temperature ( o C) Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid A B C D warm ice melt ice warm water water cools

109 25 g of 116 o C steam are bubbled into kg of water at 8 o C. Find the final temperature of the system. Calorimetry Problems 2 question #11 - [(C p, H 2 O ) (mass) (  T)] + (C v, H 2 O ) (mass) + (C p, H 2 O ) (mass) (  T) = [(C p, H 2 O ) (mass) (  T)] - [ T f ] = 997T f [q A + q B + q C ] = q D q A = [(C p, H 2 O ) (mass) (  T)] q A = [(2.042 J/g o C) (25 g) (100 o o C)] q A = J q B = (C v, H 2 O ) (mass) q B = (2256 J/g) (25 g) q B = J q C = [(C p, H 2 O ) (mass) (  T)] q C = [(4.184 J/g o C) (25 g) (T f o C)] q C = 104.5T f q D = (4.184 J/g o C) (238.4 g) (T f - 8 o C) q D = - 997T f [q A + q B + q C ] = q D T f = 997T f T f = 997T f = 1102T f 1102 T f = 68.6 o C Temperature ( o C) Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid A B C D (1000 g = 1 kg) g

110 A 322 g sample of lead (specific heat = J/g o C) is placed into 264 g of water at 25 o C. If the system's final temperature is 46 o C, what was the initial temperature of the lead? Calorimetry Problems 2 question #12 Pb T = ? o C mass = 322 g T i = 25 o C mass = 264 g LOSE heat = GAIN heat - - [(C p, Pb ) (mass) (  T)] = (C p, H 2 O ) (mass) (  T) - [(0.138 J/g o C) (322 g) (46 o C - T i )] = (4.184 J/g o C) (264 g) (46 o C- 25 o C) Drop Units: - [(44.44) (46 o C - T i )] = (1104.6) (21 o C) T i = T i = T i = 568 o C Pb T f = 46 o C

111 A sample of ice at –12 o C is placed into 68 g of water at 85 o C. If the final temperature of the system is 24 o C, what was the mass of the ice? Calorimetry Problems 2 question #13 H2OH2O T = -12 o C mass = ? g T i = 85 o C mass = 68 g GAIN heat = - LOSE heat [ q A + q B + q C ] = - [(C p, H 2 O ) (mass) (  T)] m = m = 37.8 g ice T f = 24 o C q A = [(C p, H 2 O ) (mass) (  T)] q C = [(C p, H 2 O ) (mass) (  T)] q B = (C f, H 2 O ) (mass) q A = [(2.077 J/g o C) (mass) (12 o C)] q B = (333 J/g) (mass) q C = [(4.184 J/g o C) (mass) (24 o C)] [ q A + q B + q C ] = - [(4.184 J/g o C) (68 g) (-61 o C)] 24.9 m 333 m m m q Total = q A + q B + q C 458.2

112 Endothermic Reaction Energy + Reactants  Products +  H Endothermic Reaction progress Energy Reactants Products Activation Energy

113 O Catalytic Converter C O N O C O O C O N N One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas. C O N N N O O O C O C O 2 CO(g) + 2 NO(g) N 2 (g) + 2 CO 2 (g) catalyst

114 Enthalpy Diagram H 2 O(g) H 2 O(l) H 2 (g) + ½ O 2 (g) -44 kJ Exothermic +44 kJ Endothermic  H = +242 kJ Endothermic  242 kJ Exothermic  286 kJ Endothermic  H = -286 kJ Exothermic Energy H 2 (g) + 1/2O 2 (g)  H 2 O(g) kJ  H = -242 kJ Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 211

115 Hess’s Law Calculate the enthalpy of formation of carbon dioxide from its elements. C(g) + 2O(g)  CO 2 (g) Use the following data: 2O(g)  O 2 (g)  H = kJ C(s)  C(g)  H = +720 kJ CO 2 (g)  C(s) + O 2 (g)  H = +390 kJ Smith, Smoot, Himes, pg 141 2O(g)  O 2 (g)  H = kJ C(g) + 2O(g)  CO 2 (g)  H = kJ C(g)  C(s)  H = kJ C(s) + O 2 (g)  CO 2 (g)  H = kJ

116 Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

117 Mole Pattern nose cut one (bottom) tail cut two (sides) nose cut four feet (flesh colored felt) Be sure to reverse the material when cutting two sides if a pattern is in cloth. ACTUAL SIZE

118 Fission vs. Fusion Fuse small atoms 2H 2 He NO Radioactive waste Very High Temperatures ~5,000,000 o C (SUN) Split large atoms U-235 Radioactive waste (long half-life) Nuclear Power Plants Alike Different Create Large Amounts of Energy E = mc 2 Transmutation of Elements Occurs Change Nucleus of Atoms Fusion Different Topic Fission

119 Irradiate d Spam Use fear and selective facts to promote an agenda Eating animals? Radiation = Bad Look who is funding research; it may bias the results.

120 Shielding Radiation

121 Nuclear Fission

122

123 Nuclear Power Plants map: Nuclear Energy Institute

124 Nuclear Fusion Sun + + Four hydrogen nuclei (protons) Two beta particles (electrons) One helium nucleus + Energy

125 Conservation of Mass …mass is converted into energy Hydrogen (H 2 ) H = amu Helium (He) He = amu FUSION 2 H 2  1 He + ENERGY amu x amu = amu amu This relationship was discovered by Albert Einstein E = mc 2 Energy= (mass) (speed of light) 2

126 Tokamak Reactor Fusion reactor 10,000,000 o Celcius Russian for torroidial (doughnut shaped) ring Magnetic field contains plasma

127 Cold Fusion? Fraud? Experiments must be repeatable to be valid

128 Number of half-lives Radioisotope remaining (%) Half-life of Radiation Initial amount of radioisotope t 1/2 After 1 half-lifeAfter 2 half-lives After 3 half-lives

129 Objectives - Matter Explain why mass is used as a measure of the quantity of matter. Describe the characteristics of elements, compounds, and mixtures. Solve density problems by applying an understanding of the concepts of density. Distinguish between physical and chemical properties and physical and chemical changes. Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction.

130 Objectives - Energy Identify various forms of energy. Describe changes in energy that take place during a chemical reaction. Distinguish between heat and temperature. Solve calorimetry problems. Describe the interactions that occur between electrostatic charges.

131 Law of Conservation of Energy E after = E before 2 H 2 + O 2  2 H 2 O + energy +  + WOOF!

132 Law of Conservation of Energy ENERGY CO 2 + H 2 OC 2 H 2 + O 2 PE reactants PE products KE stopper heat, light, sound E after = E before 2 H 2 + O 2  2 H 2 O + energy +  + WOOF!

133 Law of Conservation of Energy ENERGY C 2 H 2 + O 2 PE reactants PE products KE stopper heat, light, sound E after = E before 2C 2 H 2 + 5O 2  4 CO 2 + 2H 2 O + energy Energy Changes

134 First experimental image showing internal atomic structures © 2005 University of Augsburg, Experimental Physics VI,


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