《化学原理(Chemistry)英文课件》thermochemistry2.ppt

1、ThermochemistryChapter 6Energy change of chemical reaction.Thermal energy the random motion.Chemical energy the bonds of substances Nuclear energy:neutrons and protons in the atom Electrical energy:the flow of electrons Potential energy:virtue of an objects positionIt is the capacity to do work,What

2、 is the Energy?and exists in a variety of the forms.No one knows how much the exact energy is!The thermodynamics deals only with the energy changes that accompany chemical reactions.System and SurroundingsSystem is the specific part of the universe that is of interest in the study.Open systemmass&en

3、ergy exchangedclosed systemonly energy exchangedIsolated systemNothing exchanged(1)Heat is the energy transfer between a system and its surroundings that are at different temperatures.What we can measure is only the energy change!Temperature is a measure of the thermal energy.Temperature=Thermal Ene

4、rgy900C400Cgreater thermal energy?(2)Work,like heat,is the energy transfer between a system and its surroundings.Work(w)=Force(F)distance(h)=P A hhAA=P DVPressure-volume workThermodynamicsState functions determined only by the system state,regardless of how it was achieved(i.e.history).The potential

5、 energy of hiker 1 is the same as that of hiker 2.Energy change:DE=Efinal-EinitialOther changes:DP=Pfinal-PinitialDV=Vfinal-VinitialDT=Tfinal-TinitialSignificance it makes the reasoning and calculation simple.D DE =q +w q:the heat exchange with the surroundings.Attention:E is a state function,but q

6、and w are not a state function,since the latter two are only related to a process.The First Law of Thermodynamics:The First Law of Thermodynamics:The energy of the universe can be converted from one form to another,but it can not be created or destroyed.w:the work done on(or by)the system.“+”gain by

7、 the system“”lost by the systemDEsystem+DEsurroundings=0DE=Efinal-EinitialThe system energy changeExothermic process,gives off the heat to the surroundings.Endothermic process,gets the heat from the surroundings.2H2(g)+O2(g)2H2O(l)+energyH2O(g)H2O(l)+energyenergy+2HgO(s)2Hg(l)+O2(g)A sample of nitro

8、gen gas expands in volume from 1.6 L to 5.4 L at constant temperature.What is the work done in joules if the gas expands(a)against a vacuum and(b)against a constant pressure of 3.7 atm?w=-P DV(a)DV=5.4 L 1.6 L=3.8 LP=0 atmW=-0 atm x 3.8 L=0 Latm=0 joules(b)DV=5.4 L 1.6 L=3.8 LP=3.7 atmw=-3.7 atm x 3

9、.8 L=-14 Latm w=-14 Latm x 101.3 J1Latm=-1400 JD DE=q+wDV=0,w=0DE=qvAt constant volume,the heat adsorbed or released to its surrounding is totally used to increase or decrease the system internal energy.(1)At constant volume(2)At constant pressureAt constant P,DH=DE+PDVDefine enthalpy:H=E+PV H is a

10、state function,but it is not the system energy.Instead,H is only a measure of the system energy!w=-PDV,DE=qp PDVE2-E1=qp P(V2 V1)qp=(E2+PV2)(E1+PV1)qp=H2 H1=DHat constant pressure,the heat adsorbed or released is totally used to increase or decrease the system enthalpy.DE=q+wA Comparison of DH and D

11、E2Na(s)+2H2O(l)2NaOH(aq)+H2(g)DH=-367.5 kJ/molDE=DH-PDV At 25 0C and 1 atm,1 mole H2=24.5 LPDV=1 atm x 24.5 L=2.5 kJDE=-367.5 kJ/mol 2.5 kJ/mol=-370.0 kJ/mol6.4DH(Enthalpy change):H(products)H(reactants)Hproducts HreactantsDH HreactantsDH 0H(Enthalpy):describe the heat flow in a process that occurs

12、at constant pressure.Thermochemical EquationsH2O(s)H2O(l)DH=6.01 kJSystem absorbs heatEndothermicDH 0It means that:6.01 kJ are absorbed by one mole of ice that melts at 00C and 1 atm.H2O(s)H2O(l)DH=6.01 kJThermochemical Equations If the equation is reverse,the sign of DH must changeH2O(l)H2O(s)DH=-6

13、.01 kJIf both sides are mutiplied by n,DH must change by n.2H2O(s)2H2O(l)DH=2 x 6.01=12.0 kJThe physical states must be all specifiedH2O(l)H2O(g)DH=44.0 kJHow much heat is evolved when 266 g of white phosphorus(P4)burn in air?P4(s)+5O2(g)P4O10(s)DH=-3013 kJ266 g P41 mol P4123.9 g P4x3013 kJ1 mol P4x

14、=6470 kJthe absolute value of enthalpy can not be measured.i.e.DHfo=0 for any element in its most stable form.DH0(O2,g)=0fDH0(O3,g)=142 kJ/molfDH0(C,graphite)=0fDH0(C,diamond)=1.90 kJ/molfHow to make a theoretical calculation?Can we establish a reference point?DHfo Standard enthalpy of formationthe

15、heat change results when one mole of a compound is formed at 1 atm from its elements in the most stable form.C(s,graphite)+O2(g)=CO2(g)DHrxn0=-393.5 kJThen,DHf0(CO2,g)=-393.5 kJ/mol DHorxn=(H2 +H3)H1 Ca(s)O2(g)C(s,graphite)32H4 Ca(s)O2(g)12H5DHoa=H4 H1 DHoc=H2 H5 DHod=H3 H6 CaCO3(s)CaO(s)+CO2(g)H1 H

16、2H3O2(g)C(s,graphite)H6+DHob=(H5+H6)H4D DHorxn=D DHoa+D DHob+D DHoc +D DHodD DHrxn=D DH (CaO,s)+D DH (CO2,g)D DH (CaCO3,s)oofofofof=-DH (CaCO3,s)of=DH (CaO,s)of=DH (CO2,g)1atmThe standard enthalpy of reaction carried out at 1 atm.aA+bB cC+dDDH0rxndDH0(D)fcDH0(C)f=+-bDH0(B)faDH0(A)f+Benzene(C6H6)burn

17、s in air to produce carbon dioxide and liquid water.How much heat is released per mole of benzene combusted?The standard enthalpy of formation of benzene is 49.04 kJ/mol.2C6H6(l)+15O2(g)12CO2(g)+6H2O(l)=12 (393.5)+6 (187.6)2 49.04+0 =-5946 kJ-5946 kJ2 mol=-2973 kJ/mol C6H6=12 (CO2,g)+6 (H2O,l)2 (C6H

18、6,l)+15 (O2,g)DH0rxnDH0fDH0 fDH0 fDH0 fHesss Law:When reactants are converted to products,the change in enthalpy is the same regardless of the reaction taking place in one step or in a series of steps.How much is D DHof for CS2(l)given that:C(graphite)+O2(g)CO2(g)DH0 =-393.5 kJrxnS(rhombic)+O2(g)SO2

19、(g)DH0 =-296.1 kJrxnCS2(l)+3O2(g)CO2(g)+2SO2(g)DH0 =-1072 kJrxnC(graphite)+2S(rhombic)CS2(l)rxnC(graphite)+O2(g)CO2(g)DH0 =-393.5 kJ2S(rhombic)+2O2(g)2SO2(g)DH0 =-296.1x2 kJrxnCO2(g)+2SO2(g)CS2(l)+3O2(g)DH0 =+1072 kJrxn+C(graphite)+2S(rhombic)CS2(l)DH0 =-393.5+(2x-296.1)+1072=86.3 kJrxnDHsoln=Step 1+Step 2 =788 784=4 kJ/molStep 1Lattice energyStep 2hydrationDissolutionProcess of NaClWhich substance(s)could be used for melting ice?Which substance(s)could be used for a cold pack?Exercises:6.196.54 6.80 6.104 6.113