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1、Chapter 6ThermochemistrySection 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved2Capacity to do work or to produce heat.Law of conservation of energy energy can be converted from one form to another but can be neither created nor destroyed.The total energy content of the univer
2、se is constant. EnergySection 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved3Potential energy energy due to position or composition.Kinetic energy energy due to motion of the object and depends on the mass of the object and its velocity.EnergySection 6.1The Nature of EnergyCo
3、pyright Cengage Learning. All rights reserved4In the initial position, ball A has a higher potential energy than ball B.Initial PositionSection 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved5After A has rolled down the hill, the potential energy lost by A has been converted t
4、o random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B.Final PositionSection 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved6Heat involves the transfer of energy between two objects due to a temperature difference.W
5、ork force acting over a distance.Energy is a state function; work and heat are notState Function property that does not depend in any way on the systems past or future (only depends on present state).EnergySection 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved7System part of
6、the universe on which we wish to focus attention.Surroundings include everything else in the universe.Chemical EnergySection 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved8Endothermic Reaction:Heat flow is into a system.Absorb energy from the surroundings.Exothermic Reaction:
7、Energy flows out of the system.Energy gained by the surroundings must be equal to the energy lost by the system.Chemical EnergySection 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved9Is the freezing of water an endothermic or exothermic process? Explain.CONCEPT CHECK!CONCEPT C
8、HECK!Section 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved10Classify each process as exothermic or endothermic. Explain. The system is underlined in each example.a)Your hand gets cold when you touch ice.b)The ice gets warmer when you touch it.c)Water boils in a kettle being
9、heated on a stove.d)Water vapor condenses on a cold pipe.e)Ice cream melts.ExoEndoEndoExoEndoCONCEPT CHECK!CONCEPT CHECK!Section 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved11For each of the following, define a system and its surroundings and give the direction of energy tr
10、ansfer. a)Methane is burning in a Bunsen burner in a laboratory.b)Water drops, sitting on your skin after swimming, evaporate.CONCEPT CHECK!CONCEPT CHECK!Section 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved12Hydrogen gas and oxygen gas react violently to form water. Explain
11、.Which is lower in energy: a mixture of hydrogen and oxygen gases, or water?CONCEPT CHECK!CONCEPT CHECK!Section 6.1The Nature of EnergyThermodynamicsThe study of energy and its interconversions is called thermodynamics.Law of conservation of energy is often called the first law of thermodynamics.Cop
12、yright Cengage Learning. All rights reserved13Section 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved14Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system.To change the internal energy of a system:E = q + wq repre
13、sents heatw represents workInternal EnergySection 6.1The Nature of EnergyWork vs Energy FlowCopyright Cengage Learning. All rights reservedSection 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved16Thermodynamic quantities consist of two parts:Number gives the magnitude of the c
14、hange.Sign indicates the direction of the flow. Internal EnergySection 6.1The Nature of EnergyInternal EnergySign reflects the systems point of view.Endothermic Process:q is positiveExothermic Process:q is negativeCopyright Cengage Learning. All rights reserved17Section 6.1The Nature of EnergyCopyri
15、ght Cengage Learning. All rights reserved18Sign reflects the systems point of view.System does work on surroundings:w is negativeSurroundings do work on the system:w is positiveInternal EnergySection 6.1The Nature of Energy19Work = P A h = PVP is pressure.A is area.h is the piston moving a distance.
16、V is the change in volume.WorkSection 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved20For an expanding gas, V is a positive quantity because the volume is increasing. Thus V and w must have opposite signs:w = PV To convert between Latm and Joules, use 1 Latm = 101.3 J.WorkSec
17、tion 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved21Which of the following performs more work?a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L.b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L. They perform the same amount of work.EXERCISE!
18、EXERCISE!Section 6.1The Nature of EnergyCopyright Cengage Learning. All rights reserved22Determine the sign of E for each of the following with the listed conditions:a)An endothermic process that performs work. |work| |heat| |work| |heat| |work| |heat| E = negative E = positive E = positive E = nega
19、tiveCONCEPT CHECK!CONCEPT CHECK!Section 6.2Enthalpy and CalorimetryChange in EnthalpyState functionH = q at constant pressureH = Hproducts HreactantsCopyright Cengage Learning. All rights reserved23Section 6.2Enthalpy and CalorimetryConsider the combustion of propane:C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(
20、l)H = 2221 kJAssume that all of the heat comes from the combustion of propane. Calculate H in which 5.00 g of propane is burned in excess oxygen at constant pressure. 252 kJCopyright Cengage Learning. All rights reserved24EXERCISE!EXERCISE!Section 6.2Enthalpy and CalorimetryCalorimetryScience of mea
21、suring heatSpecific heat capacity:The energy required to raise the temperature of one gram of a substance by one degree Celsius.Molar heat capacity:The energy required to raise the temperature of one mole of substance by one degree Celsius.Copyright Cengage Learning. All rights reserved25Section 6.2
22、Enthalpy and CalorimetryCalorimetryIf two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic.An endothermic reaction cools the solution.Copyright Cengage Learning. All rights reserved26Section 6.2Enthalpy and Calorim
23、etryA CoffeeCup Calorimeter Made of Two Styrofoam CupsCopyright Cengage Learning. All rights reserved27Section 6.2Enthalpy and CalorimetryCalorimetryEnergy released (heat) = s m Ts = specific heat capacity (J/Cg)m = mass of solution (g)T = change in temperature (C)Copyright Cengage Learning. All rig
24、hts reserved28Section 6.2Enthalpy and CalorimetryA 100.0 g sample of water at 90C is added to a 100.0 g sample of water at 10C.The final temperature of the water is:a) Between 50C and 90Cb) 50Cc) Between 10C and 50CCopyright Cengage Learning. All rights reserved29CONCEPT CHECK!CONCEPT CHECK!Section
25、6.2Enthalpy and CalorimetryA 100.0 g sample of water at 90.C is added to a 500.0 g sample of water at 10.C.The final temperature of the water is:a) Between 50C and 90Cb) 50Cc) Between 10C and 50CCalculate the final temperature of the water. 23CCopyright Cengage Learning. All rights reserved30CONCEPT
26、 CHECK!CONCEPT CHECK!Section 6.2Enthalpy and CalorimetryYou have a Styrofoam cup with 50.0 g of water at 10.C. You add a 50.0 g iron ball at 90. C to the water. (sH2O = 4.18 J/Cg and sFe = 0.45 J/Cg)The final temperature of the water is:a) Between 50C and 90Cb) 50Cc) Between 10C and 50CCalculate the
27、 final temperature of the water.18CCopyright Cengage Learning. All rights reserved31CONCEPT CHECK!CONCEPT CHECK!Section 6.3Hesss LawIn going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a ser
28、ies of steps.Copyright Cengage Learning. All rights reserved32Section 6.3Hesss LawThis reaction also can be carried out in two distinct steps, with enthalpy changes designated by H2 and H3. N2(g) + O2(g) 2NO(g) H2 = 180 kJ2NO(g) + O2(g) 2NO2(g) H3 = 112 kJ N2(g) + 2O2(g) 2NO2(g) H2 + H3 = 68 kJH1 =
29、H2 + H3 = 68 kJN2(g) + 2O2(g) 2NO2(g) H1 = 68 kJCopyright Cengage Learning. All rights reserved33Section 6.3Hesss LawThe Principle of Hesss LawCopyright Cengage Learning. All rights reserved34Section 6.3Hesss LawCopyright Cengage Learning. All rights reserved35Section 6.3Hesss LawCharacteristics of
30、Enthalpy ChangesIf a reaction is reversed, the sign of H is also reversed.The magnitude of H is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of H is multiplied by the same integer.C
31、opyright Cengage Learning. All rights reserved36Section 6.3Hesss LawExampleConsider the following data:Calculate H for the reactionCopyright Cengage Learning. All rights reserved37Section 6.3Hesss LawProblem-Solving StrategyWork backward from the required reaction, using the reactants and products t
32、o decide how to manipulate the other given reactions at your disposal.Reverse any reactions as needed to give the required reactants and products.Multiply reactions to give the correct numbers of reactants and products.Copyright Cengage Learning. All rights reserved38Section 6.3Hesss LawExampleRever
33、se the two reactions:Desired reaction:Copyright Cengage Learning. All rights reserved39Section 6.3Hesss LawExampleMultiply reactions to give the correct numbers of reactants and products: 4( ) 4( ) 3( ) 3( )Desired reaction:Section 6.3Hesss LawExampleFinal reactions:Desired reaction: H = +1268 kJ Co
34、pyright Cengage Learning. All rights reserved41Section 6.4Standard Enthalpies of FormationStandard Enthalpy of Formation (fH)Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states.Copyright Cengage Learning. All righ
35、ts reserved42Section 6.4Standard Enthalpies of FormationConventional Definitions of Standard StatesFor a CompoundFor a gas, pressure is exactly 1 atm.For a solution, concentration is exactly 1 M.Pure substance (liquid or solid)For an ElementThe form N2(g), K(s) in which it exists at 1 atm and 25C.Co
36、pyright Cengage Learning. All rights reserved43Section 6.4Standard Enthalpies of FormationA Schematic Diagram of the Energy Changes for the Reaction CH4(g) + 2O2(g) CO2(g) + 2H2O(l)reactionH = (75 kJ) + 0 + (394 kJ) + (572 kJ) = 891 kJCopyright Cengage Learning. All rights reserved44Section 6.4Stand
37、ard Enthalpies of FormationProblem-Solving Strategy: Enthalpy Calculations1.When a reaction is reversed, the magnitude of H remains the same, but its sign changes.2.When the balanced equation for a reaction is multiplied by an integer, the value of H for that reaction must be multiplied by the same
38、integer.Copyright Cengage Learning. All rights reserved45Section 6.4Standard Enthalpies of FormationProblem-Solving Strategy: Enthalpy Calculations3.The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products:Hrxn = npHf(products) - nr
39、Hf(reactants)4. Elements in their standard states are not included in the Hreaction calculations because Hf for an element in its standard state is zero.Copyright Cengage Learning. All rights reserved46Section 6.4Standard Enthalpies of FormationCalculate H for the following reaction:2Na(s) + 2H2O(l)
40、 2NaOH(aq) + H2(g)Given the following information: Hf (kJ/mol) Na(s)0 H2O(l) 286 NaOH(aq) 470 H2(g)0H = 368 kJCopyright Cengage Learning. All rights reserved47EXERCISE!EXERCISE!Section 6.5Present Sources of EnergyFossil FuelsPetroleum, Natural Gas, and CoalWoodHydroNuclearCopyright Cengage Learning.
41、 All rights reserved48Section 6.5Present Sources of EnergyEnergy Sources Used in the United StatesCopyright Cengage Learning. All rights reserved49Section 6.5Present Sources of EnergyThe Earths AtmosphereTransparent to visible light from the sun.Visible light strikes the Earth, and part of it is cha
42、nged to infrared radiation.Infrared radiation from Earths surface is strongly absorbed by CO2, H2O, and other molecules present in smaller amounts in atmosphere.Atmosphere traps some of the energy and keeps the Earth warmer than it would otherwise be.Copyright Cengage Learning. All rights reserved50Section 6.5Present Sources of EnergyThe Earths AtmosphereCopyright Cengage Learning. All rights reserved51Section 6.6New Energy SourcesCoal ConversionHydrogen as a FuelOther Energy AlternativesOil shaleEthanolMethanolSeed oilCopyright Cengage Learning. All rights reserved52
