無(wú)機(jī)化學(xué)課件：11 Ch10 Ch 18 Elements of Chemical Thermodynamics

1、Chemistry: Atoms FirstJulia Burdge & Jason OverbyChapter 10Energy Changes in Chemical ReactionsThermochemistry1010.1 Energy and Energy Changes10.2 Introduction to ThermodynamicsStates and State FunctionsThe First Law of ThermodynamicsWork and Heat10.3 EnthalpyReactions Carried Out at Constant Volume

2、 or at Constant PressureEnthalpy and Enthalpy ChangesThermochemical Equations10.4 CalorimetrySpecific Heat and Heat CapacityConstant-Pressure CalorimetryConstant-Volume Calorimetry10.5 Hesss Law10.6 Standard Enthalpies of Formation10.7 Bond Enthalpy and the Stability of Covalent Molecules10.8 Lattic

3、e Energy and the Stability of Ionic CompoundsThe Born-Haber CycleComparison of Ionic and Covalent CompoundsEnergy and Energy ChangesThe system is a part of the universe that is of specific interest.The surroundings constitute the rest of the universe outside the system.The system is usually defined

4、as the substances involved in chemical and physical changes.SystemSurroundingsUniverse = System + Surroundings10.1Energy and Energy ChangesThermochemistry is the study of heat (the transfer of thermal energy) in chemical reactions.Heat is the transfer of thermal energy.Heat is either absorbed or rel

5、eased during a process.SurroundingsheatEnergy and Energy ChangesAn exothermic process occurs when heat is transferred from the system to the surroundings.“Feels hot!”SurroundingsUniverse = System + SurroundingsheatSystem2H2(g) + O2(g) 2H2O(l) + energyEnergy and Energy ChangesAn endothermic process o

6、ccurs when heat is transferred from the surroundings to the system.“Feels cold”energy + 2HgO(s) 2Hg(l) + O2(g)SurroundingsUniverse = System + SurroundingsheatSystemIntroduction to ThermodynamicsThermodynamics is the study of the interconversion of heat and other kinds of energy.In thermodynamics, th

7、ere are three types of systems:An open system can exchange mass and energy with the surroundings.A closed system allows the transferof energy but not mass.An isolated system does not exchangeeither mass or energy with its surroundings.10.2openmass & energyExchange:closedenergyisolatednothingInternal

8、 energyInternal energy, also called thermodynamic energythe energy associated with the random, disordered motion of molecules or particles.It is separated in scale from the macroscopic ordered energy.Its absolute value is unknown.Its symbol and unit are U and joule respectively.Internal EnergyIntern

9、al energy: 2 components Kinetic energy - molecular motionPotential energy - attractive/repulsive interactionsInternal Energy We cannot calculate the exact value of internal energy with any certaintyWe can calculate the changes in energy of the system experimentallyU = U(products) - U(reactants)Intro

10、duction to ThermodynamicsState functions are properties that are determined by the state of the system, regardless of how that condition was achieved.The magnitude of change depends only on the initial and final states of the system.EnergyPressureVolumeTemperatureState functions are properties that

11、are determined by the state of the system, regardless of how that condition was achieved.Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.energy, pressure, volume, temperatureDU = Ufinal - UinitialDP = Pfinal - PinitialDV = Vfinal - VinitialDT = Tfinal - Tini

12、tialThe First Law of ThermodynamicsThe first law of thermodynamics states that energy can be converted from one form to another, but cannot be created or destroyed.U is the change in the internal energy.“sys” and “surr” denote system and surroundings, respectively.U = Uf Ui; the difference in the en

13、ergies of the initial and final states.Usys + Usurr = 0 Usys = UsurrWork and HeatThe overall change in the systems internal energy is given by:q is heatq is positive for an endothermic process (heat absorbed by the system)q is negative for an exothermic process (heat released by the system)w is work

14、w is positive for work done on the systemw is negative for work done by the systemU = q + wWork and HeatU = q + wThe Sign Conventions* for q, w and DUqw+=DU+-+-depends on sizes of q and wdepends on sizes of q and w* For q: + means system gains heat; - means system loses heat.* For w: + means work do

15、ne on system; - means work done by system. Worked Example 10.1Strategy Combine the two contributions to internal energy using U = q + w and the sign conventions for q and w.Calculate the overall change in internal energy, U, (in joules) for a system that absorbs 188 J of heat and does 141 J of work

16、on its surroundings.Solution The system absorbs heat, so q is positive. The system does work on the surroundings, so w is negative.U = q + w = 188 J + (-141 J) = 47 JThink About It Consult Table 10.1 to make sure you have used the proper sign conventions for q and w.EnthalpySodium azide detonates to

17、 give a large quantity of nitrogen gas.Under constant volume conditions, pressure increases:2NaN3(s) 2Na(s) + 3N2(g)10.3EnthalpySodium azide detonates to give a large quantity of nitrogen gas.Under constant volume conditions, pressure increases:2NaN3(s) 2Na(s) + 3N2(g)EnthalpyPressure-volume, or PV

18、work, is done when there is a volume change under constant pressure.w = PVP is the external opposing pressure.V is the change in the volume of the container. Worked Example 10.2Strategy Determine change in volume (V), identify the external pressure (P), and use w = PV to calculate w. The result will

19、 be in Latm; use the equality 1 Latm = 101.3 J to convert to joules.Determine the work done (in joules) when a sample of gas extends from 552 mL to 891 mL at constant temperature (a) against a constant pressure of 1.25 atm, (b) against a constant pressure of 1.00 atm, and (c) against a vacuum (1 Lat

20、m = 101.3 J).Solution V = (891 552)mL = 339 mL. (a) P = 1.25 atm, (b) P = 1.00 atm, (c) P = 0 atm.(a) w = -(1.25 atm)(339 mL)(b) w = -(1.00 atm)(339 mL)1 L1000 mL101.3 J1 Latm= -42.9 J1 L1000 mL101.3 J1 Latm= -34.3 J Worked Example 10.2 (cont.)Solution(c) w = -(0 atm)(339 mL)1 L1000 mL101.3 J1 Latm=

21、 0 JThink About It Remember that the negative sign in the answers to part (a) and (b) indicate that the system does work on the surroundings. When an expansion happens against a vacuum, no work is done. This example illustrates that work is not a state function. For an equivalent change in volume, t

22、he work varies depending on external pressure against which the expansion must occur.EnthalpyPressure-volume, or PV work, is done when there is a volume change under constant pressure.w = PVWhen a change occurs at constant volume, V = 0 and no work is done.U = q + w qV = UU = q PV substituteEnthalpy

23、Under conditions of constant pressure:U = q + w qP = U + PVU = q PV EnthalpyThe thermodynamic function of a system called enthalpy (H) is defined by the equation:H = U + PVA note about SI units:Pressure:pascal; 1Pa = 1 kg/(m . s2)Volume:cubic meters; m3PV:1kg/(m . s2) x m3 = 1(kg . m2)/s2 = 1 JEntha

24、lpy:joulesU, P, V, and H are all state functions.Enthalpy and Enthalpy ChangesFor any process, the change in enthalpy is:H = U + (PV)(1)H = U + PVIf pressure is constant:(2)U = H + PVRearrange to solve for U:(3)qp = U + PVRemember, qp:(4)qp = (H PV) + PVSubstitute equation (3) into equation (4) and

25、solve:(5)qp = Hfor a constant-pressure processEnthalpy and Enthalpy ChangesThe enthalpy of reaction (Hr) is the difference between the enthalpies of the products and the enthalpies of the reactants:Assumes reactions in the lab occur at constant pressureH 0 (positive)endothermic processH 0endothermic

26、H2O(s) H2O(l) H = +6.01 kJ/molThermochemical EquationsConcepts to consider:Is this a constant pressure process?What is the system?What are the surroundings?H 0C; H = -6.01 kJ/molEntropy18.2To predict spontaneity, both the enthalpy and entropy must be known.Entropy (S) of a system is a measure of how

27、 spread out or how dispersed the systems energy is. EntropySpontaneity is favored by an increase in entropy.k is the Boltzmann constant (1.38 x 1023 J/K)W is the number of different arrangementsS = k ln WThe number of arrangements possible is given by:X is the number of cells in a volumeN is the num

28、ber of moleculesW = X NEntropyEntropyThere are three possible states for this system:1)One molecule on each side (eight possible arrangements)2)Both molecules on the left (four possible arrangements)3)Both molecules on the right (four possible arrangements)The most probable state has the largest num

29、ber of arrangements.Entropy Changes in a SystemThe change in entropy for a system is the difference in entropy of the final state and the entropy of the initial state.Alternatively:18.3Ssys = Sfinal Sinitial Worked Example 18.1Strategy This is the isothermal expansion of an ideal gas. Because the mo

30、lecules spread out to occupy a greater volume, we expect there to be an increase in the entropy of the system. Use Ssys = Sfinal Sinitial to solve for Ssys. R = 8.314 J/Kmol, n = 1.0 mole, Vfinal = 5.0 L, and Vinitial = 2.5 L.Determine the change in entropy for 1.0 mole of an ideal gas originally co

31、nfined to one-half of a 5.0-L container when the gas is allowed to expand to fill the entire container at constant temperature.SolutionSsys = nR lnVfinalVinitial= 1.0 mol 8.314 JK molln5.0 L2.5 L= 5.8 J/KThink About It Remember that for a process to be spontaneous, something must favor spontaneity.

32、If the process is spontaneous but not exothermic (in this case, there is no enthalpy change), then we should expect Ssys to be positive. Worked Example 18.1 (cont.)Solution These equilibrium concentrations are then substituted into the equilibrium expression to giveBecause we expect x to be very sma

33、ll (even smaller than 1.3410-3 Msee above), because the ionization of CH3COOH is suppressed by the presence of CH3COO-, we assume(0.10 x) M 0.10 Mand(0.050 + x) M 0.050 MTherefore, the equilibrium expression simplifies toand x = 3.610-5 M. According to the equilibrium table, H+ = x, so pH = log(3.61

34、0-5) = 10-5 =(x)(0.050 + x)0.010 x1.810-5 =(x)(0.050)0.010Entropy Changes in a SystemThe standard entropy (S) is the absolute entropy of a substance at 1 atm.Temperature is not part of the standard state definition and must be specified.Entropy Changes in a SystemThere are several important

35、trends in entropy: Sliquid Ssolid Sgas Sliquid S increases with molar mass S increases with molecular complexity S increases with the mobility of a phase (for an element with two or more allotropes)Entropy Changes in a SystemIn addition to translational motion, molecules exhibit vibrations and rotat

36、ions.For a chemical reactionaA + bB cC + dDAlternatively, Entropy Changes in a SystemrxnS = cS (C) + dS (D) aS (A) + bS (B)rxnS = nS (products) mS (reactants) Worked Example 18.2Strategy Look up standard enthalpy values and calculate rxnS. Just as we did when we calculated standard enthalpies of rea

37、ction, we consider stoichiometric coefficients to be dimensionlessgiving rxnS units of J/Kmol.From Appendix 2, S CaCO3(s) = 92.9 J/Kmol, S CaO(s) = 39.8 J/Kmol, S CO2(g) = 213.6 J/Kmol, S N2(g) = 191.5 J/Kmol, S H2(g) = 131.0 J/Kmol, S NH3(g) = 193.0 J/Kmol, S Cl2(g) = 223.0 J/Kmol, and S HCl(g) = 1

38、87.0 J/Kmol.From the standard enthalpy values in Appendix 2, calculate the standard entropy changes for the following reactions at 25C.(a) CaCO3(s) CaO(s) + CO2(g)(b) N2(g) + 3H2(g) 2NH3(g)(c) H2(g) + Cl2(g) 2HCl(g) Worked Example 18.2 (cont.)Solution (a) Srxn = S (CaO) + S (CO2) S (CaCO3)= (39.8 J/

39、Kmol) + (213.6 J/Kmol) (92.9 J/Kmol)= 160.5 J/Kmol(b) Srxn = 2S (NH3) S (N2) + 3S (H2)=(2)(193.0 J/Kmol) (191.5 J/Kmol) + (3)(131.0 J/Kmol)= 198.5 J/Kmol(c) Srxn = 2S (HCl) S (H2) + S(Cl2)= (2)(187.0 J/Kmol) (131.0 J/Kmol) + (223.0 J/Kmol)= 20.0 J/KmolThink About It Remember to multiply each standar

40、d entropy value by the correct stoichiometric coefficient. Like Equation 10.18, Equation 18.5 can only be used with a balanced chemical equation.Entropy Changes in a SystemSeveral processes that lead to an increase in entropy are: Melting Vaporization or sublimation Temperature increase Reaction res

41、ulting in a greater number of gas moleculesEntropy Changes in a SystemThe process of dissolving a substance can lead to either an increase or a decrease in entropy, depending on the nature of the solute.Molecular solutes (i.e. sugar): entropy increasesIonic compounds: entropy could decrease or incre

42、ase Worked Example 18.3Strategy Consider the change in energy/mobility of atoms and the resulting change in number of possible positions that each particle can occupy in each case. An increase in the number of arrangements corresponds to an increase in entropy and therefore a positive S.For each pro

43、cess, determine the sign of S for the system: (a) decomposition of CaCO3(s) to give CaO(s) and CO2(g), (b) heating bromine vapor from 45C to 80C, (c) condensation of water vapor on a cold surface, (d) reaction of NH3(g) and HCl(g) to give NH4Cl(s), and (e) dissolution of sugar in water.Solution Incr

44、eases in entropy generally accompany solid-to-liquid, liquid-to-gas, and solid-to-gas transitions; the dissolving of one substance in another; a temperature increase; and reactions that increase the net number of moles of gas.S is (a) positive(b) positive(c) negative(d) negative(e) positiveThink Abo

45、ut It For reactions involving only liquids and solids, predicting the sign of S can be more difficult, but in many such cases an increase in the total number of molecules and/or ions is accompanied by an increase of entropy.Entropy Changes in the UniverseCorrectly predicting the spontaneity of a pro

46、cess requires us to consider entropy changes in both the system and the surroundings.An ice cube spontaneously melts in a room at 25C.A cup of hot water spontaneously cools to room temperature.The entropy of both the system AND surroundings are important!18.4PerspectiveComponentsSSystemicepositiveSu

47、rroundingseverything elsenegativePerspectiveComponentsSSystemhot waternegativeSurroundingseverything elsepositiveEntropy Changes in the UniverseThe change in entropy of the surroundings is directly proportional to the enthalpy of the system.The second law of thermodynamics states that for a process

48、to be spontaneous, Suniverse must be positive.Suniverse = Ssys + Ssurr Entropy Changes in the UniverseThe second law of thermodynamics states that for a process to be spontaneous, Suniverse must be positive.Suniverse 0 for a spontaneous process Suniverse 0 for a nonspontaneous processSuniverse = 0 f

49、or an equilibrium process Suniverse = Ssys + Ssurr Worked Example 18.4Strategy For each process, use Srxn = nS(products) mS(reactants) to determine S sys; Ssurr = (Hsys/T)and H sys = nH f(products) mH f(reactants) to determine H sys and S surr. At the specified temperature, the process is spontaneou

50、s if S sys and S surr sum to a positive number, nonspontaneous is they sum to a negative number, and an equilibrium process if they sum to zero. Note that because the reaction is the system, Srxn and Ssys are used interchangeably.Determine if each of the following is a spontaneous process, a nonspon

51、taneous process, or an equilibrium process at the specified temperature: (a) H2(g) + I2(g) 2HI(g) at 0C, (b) CaCO3(s) CaO(s) + CO2(g) at 200C, (c) CaCO3(s) CaO(s) + CO2(g) at 1000C, (d) Na(s) Na(l) at 98C. (Assume that the thermodynamic data in Appendix 2 do not vary with temperature.) Worked Exampl

52、e 18.4 (cont.)Solution (a) S H2(g) = 131.0 J/Kmol, S I2(g) = 260.57 J/Kmol, S HI(g) = 206.3 J/Kmol; fH H2(g) = 0 kJ/mol, fH I2(g) = 62.25 kJ/mol, fH HI(g) = 25.9 kJ/mol.rxnS = 2S (HI) S (H2) + S (I2) = (2)(206.3 J/Kmol) 131.0 J/Kmol + 160.57 J/Kmol = 21.03 J/KmolrxnH = 2 fH (HI) fH (H2) + fH (I2) =

53、(2)(25.9 kJ/mol) 0 kJ/mol + 62.26 kJ/mol = 10.5 kJ/molsurrS = = = 0.0385 kJ/Kmol = 38.5 J/KmolSuniverse = Ssys + Ssurr = 21.03 J/Kmol + 38.5 J/Kmol = 59.5 J/KmolSuniverse is positive; therefore the reaction is spontaneous at 0C.(10.5 kJ/mol)273 KrxnHT Worked Example 18.4 (cont.)Solution (b) In Worke

54、d Example 18.2(a), we determined that for this reaction, rxnS = 160.5 J/Kmol; fH CaCO3(s) = 1206.9 kJ/mol, fH CaO(s) = 635.6 kJ/mol, f H CO2(g) = 393.5 kJ/mol.(b), (c)rxnS = 160.5 J/KmolrxnH = fH (CaO) + fH (CO2) fH (CaCO3) = -635.6 kJ/mol + (393.5 kJ/mol) (1206.9 kJ/mol) = 177.8 kJ/mol(b) T = 200C

55、andSsurr = = = 0.376 kJ/Kmol = 376 J/KmolSuniverse = Ssys + Ssurr = 160.5 J/Kmol + (376 J/Kmol) = 216 J/Kmolsuniverse is negative, therefore the reaction is nonspontaneous at 200C. (177.8 kJ/mol)473 KHsysT Worked Example 18.4 (cont.)Solution (c) T = 1000C andSsurr = = = 0.1397 kJ/Kmol = 139.7 J/Kmol

56、Suniverse = Ssys + Ssurr = 160.5 J/Kmol + (139.7 J/Kmol) = 20.8 J/KmolIn this case, Suniverse is positive; therefore, the reaction is spontaneous at 1000C.(177.8 kJ/mol)473 KHsysT Worked Example 18.4 (cont.)Solution (d) S Na(s) = 51.05 J/Kmol, S Na(l) = 57.56 J/Kmol; fH Na(s) = 0 kJ/mol, fH Na(l) =

57、2.41 kJ/mol.rxnS = S Na(l) S Na(s) = 57.56 J/Kmol 51.05 J/Kmol = 6.51 J/KmolrxnH = fH Na(l) fH Na(s) = 2.41 kJ/mol 0 kJ/mol = 2.41 kJ/molSsurr = = = 0.0650 kJ/Kmol = 6.50 J/KmolSuniverse = Ssys + Ssurr = 6.51 J/Kmol 6.50 J/Kmol = 0.01 J/Kmol 0Suniverse is zero; therefore, the reaction is an equilibr

58、ium process at 98C. In fact, this is the melting point of sodium.(2.41 kJ/mol)371 KHrxnT Think About It Remember that standard enthalpies of formation have units of kJ/mol, whereas standard absolute entropies have units of J/Kmol. Make sure that you convert kilojoules to joules, or vice versa, befor

59、e combining the terms.Entropy Changes in the UniverseThe third law of thermodynamics states that the entropy of a perfect crystalline substance is zero at absolute zero.Entropy increases in a substance as temperature increases from absolute zero.Predicting SpontaneityMeasurements on the surroundings

60、 are seldom made, limiting the use of the second law of thermodynamics.Gibbs free energy (G) or simply free energy can be used to express spontaneity more directly.The change in free energy for a system is:18.5G = H TSG = H TS Predicting SpontaneityUsing the Gibbs free energy, it is possible to make