Entropy, Free Energy, and Equilibrium熵自由能與平衡ppt課件

1、ThermodynamicsEntropy, Free Energy, and EquilibriumChapter 183/eCopyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display._ Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 00C and ice

2、 melts above 00C Heat flows from a hotter object to a colder object A gas expands into an evacuated bulb Iron exposed to oxygen and water forms rustspontaneousnonspontaneous18.218.2Refresh your memory: What is enthalpy?Enthalpy is a thermodynamic property that describes heat changes taking place at

3、constant _.Enthalpy is an _propertyits magnitude depends on the amount of the substance present.Enthalpy cannot be determined, so we measure H, the change in enthalpy.In earlier chapters, we examined Hrxn, Hsoln, and HEnthalpy is a _functionfIdentify the following reactions as exothermic or endother

4、mic. Does a decrease in enthalpy mean a reaction proceeds spontaneously? CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = - 890.4 kJH+ (aq) + OH- (aq) H2O (l) DH0 = - 56.2 kJH2O (s) H2O (l) DH0 = 6.01 kJNH4NO3 (s) NH4+(aq) + NO3- (aq) DH0 = 25 kJH2OThese are all reactions:18.2So we see that a decrease in

5、enthalpy does not mean a reaction proceeds spontaneously.What do we need to know in order to predict if a reaction is spontaneous? Two things (p.577)._AND_ The greater the disorder or randomness of a system, the _ _ _.The more ordered a system is, the less its _.Entropy (S) is a measure of the _ of

6、a system.orderSdisorderSDS = Sf - SiIf the change from initial to final results in an increase in randomnessSf SiDS 0For any substance, the _ state is more ordered than the _ state, and the _ state is more ordered than _ stateSsolid Sliquid 018.2Thermodynamics_ _ are properties that are determined b

7、y 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, enthalpy, pressure, volume, temperature18.7, entropyStandard Entropy Values (S ) for Some Substances at 25CProcesses that lead to

8、 an _ in entropy (DS 0)18.2p.579Look at page 580, Example 18.1.Lets work through it together.Then, lets look at the Practice Exercise below it.How does the entropy of a system change for each of the following processes?(a) Condensing water vaporRandomness _(b) Forming sucrose crystals from a supersa

9、turated solutionRandomness _(c) Heating hydrogen gas from 600C to 800CRandomness _(d) Subliming dry iceRandomness _18.2Ex.18.1, p.580Entropy _Entropy _Entropy _Entropy _First Law of ThermodynamicsEnergy can be converted from one form to another, but energy cannot be created or destroyed.Second Law o

10、f ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.DSuniv = DSsys + DSsurr 0Spontaneous process:DSuniv = DSsys + DSsurr = 0Equilibrium process:18.3Entropy Changes in the System (DSsys)aA + bB cC + dDDS0rxndS0(D)cS0(C)= +bS0(

11、B)aS0(A)+DS0rxnnS0(products)= SmS0(reactants)SThe _ _ of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C.rxn18.3Sn = the sum of the stoichiometric coefficients of the productsSm = the sum of the stoichiometric coefficients of the reactantsWhat does the superscripte

12、d zero mean?Look at the bottom of page 581 and top of page 582, Example 18.2.Lets work through it together.Then, lets look at the Practice Exercise below it.Entropy Changes in the System (DSsys)18.3What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g)S0(CO

13、) = 197.9 J/KmolS0(O2) = 205.0 J/KmolS0(CO2) = 213.6 J/KmolDS0rxn= 2 x S0(CO2) 2 x S0(CO) + S0 (O2) DS0rxn= 427.2 395.8 + 205.0 = -173.6 J/Kmol Where do we get these values?You should work through the other parts of this Practice Exercise.Entropy Changes in the System (DSsys)18.3When gases are produ

14、ced or consumed If a reaction produces more gas molecules than it consumes, DS0 0.If the total number of gas molecules diminishes, DS0 0Endothermic ProcessDSsurr 0Spontaneous process:TDSuniv = DHsys + TDSsys 0Multiply by TBut we can predict spontaneity more directly.18.4Substituting DHsys/T for DSsu

15、rrDSuniv = DSsys DHsys 0TMultiply by 1TDSuniv = DHsys TDSsys 0p.586This says that for a process carried out at constant P and T, if the changes in enthalpy (H) and entropy (S) of the system are such that DHsys TDSsys is less than zero, then the process must be spontaneous.Gibbs Free Energy (G)For a

16、constant-temperature process:_The change in Gibbs free energy (DG)18.4If DG is negative, there is a release of usable energy, and the reaction is spontaneous!If DG is positive, the reaction is not spontaneous!G = H TSAll quantities in the above equation refer to the systemFor a constant-temperature

17、process:DG = DHsys TDSsysDG 0 The reaction is _ as written. The reaction is _ in the reverse directionDG = 0 The reaction is _ .18.418.4aA + bB cC + dDDG0rxndDG0 (D)fcDG0 (C)f= + -bDG0 (B)faDG0 (A)f+DG0rxnnDG0 (products)f= SmDG0 (reactants)fS-The _ is the free-energy change for a reaction when it oc

18、curs under standard-state conditions.rxnSn = the sum of the stoichiometric coefficients of the productsSm = the sum of the stoichiometric coefficients of the reactantsThe superscripted zero means _18.4_(DG0) is the free-energy change that occurs when: * 1 mole of the compound * is formed from its el

19、ements * in their standard states.fDG0 of any element in _ is zero.fConventions for Standard StatesfLook at page 587f, Example 18.4.Lets work through it together.Then, lets look at the Practice Exercise below it.2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0rxnnDG0 (products)f= SmDG0 (reactants)fS-Wha

20、t is the standard free-energy change for the following reaction at 25 0C?DG0rxn6DG0 (H2O)f12DG0 (CO2)f=+ - 2DG0 (C6H6)fDG0rxn= 12x394.4 + 6x237.2 2x124.5 = -6405 kJIs this reaction spontaneous at 25 0C?18.4Practice Exercise 18.4 (b), p.5882C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0rxnnDG0 (products

21、)f= SmDG0 (reactants)fS-What is the standard free-energy change for the following reaction at 25 0C?DG0rxn6DG0 (H2O)f12DG0 (CO2)f=+ - 2DG0 (C6H6)fDG0rxn= 12x394.4 + 6x237.2 2x124.5 = -6405 kJIs this reaction spontaneous at 25 0C?DG0 = -6405 kJ 0This reaction is spontaneous18.4DG = DH TDSIf both DH a

22、nd DS are positive, then DG will be negative only when the TDS term is larger than DH. This occurs only when T is large.If DH is positive and DS is negative, DG will always be positiveregardless of the temperature.If DH is negative and DS is positive, then DG will always be negative regardless of te

23、mperature.If DH is negative and DS is negative, then DG will be negative only when TDS is smaller in magnitude than DH. This condition is met when T is small.p.588DG = DH TDS18.4p.589Factors Affecting the Sign of DG in the Relationship DG = DH TDSCaCO3 (s) CaO (s) + CO2 (g)Temperature and Spontaneit

24、y of Chemical ReactionsAt what temperature will this reaction become spontaneous? Orat what temperature does DG become zeroor negative?1. Use Eqn 6.16 to calculate DH2. Use Eqn 18.4 to calculate DS3. Substitute these into DG = DH TDSto calculate DG (130.0 kJ)4. Since this is a large positive number,

25、 we set DG = 0 and solve the equation for TFollow along on pp.589fCaCO3 (s) CaO (s) + CO2 (g)Temperature and Spontaneity of Chemical Reactions“Two points are worth making.” Read quote p.590.First, we used DH and DS values at 25C to calculate changes that occur at much higher temperatures. Both DH an

26、d DS change with temperature, so the DG value will not be accuratebut it is a helpful, ballpark estimate.Second, dont think nothing happens below 835C, and then suddenly, at 835C, CaCO3 begins to decompose. Look at the graph on the next frame to see what happens.CaCO3 (s) CaO (s) + CO2 (g)DH0 = 177.8 kJDS0 = 160.5 J/KDG0 = DH0 TDS0At 25 0C, DG0 = 130.0 kJDG0 = 0 at 835 0C18.4Temperature and Spontaneity of Chemical Reactionsp.590The condition DG = 0 applies to any phase transit