Entropy-and-Free-Energy熵與自由能-PPT精選課件

1、Entropy and Free Energy Spontaneous vs. non-spontaneous thermodynamics vs. kinetics entropy = randomness (So) Gibbs free energy (Go) Go for reactions - predicting spontaneous direction thermodynamics of coupled reactions Grxn versus Gorxn predicting equilibrium constants from Gorxn 1Entropy and Free

2、 EnergyHow can we predict if a reaction can occur, given enough time?Note: Thermodynamics DOES NOT say how quickly (or slowly) a reaction will occur. To predict if a reaction can occur at a reasonable rate, one needs to consider: Some processes are spontaneous; others never occur. WHY ?THERMODYNAMIC

3、SKINETICS2Product-Favored Reactionse.g. thermite reactionFe2O3(s) + 2 Al(s) 2 Fe(s) + Al2O3(s)DH = - 848 kJIn general, product-favored reactions are exothermic.3Non-exothermic spontaneous reactionsBut many spontaneous reactions or processes are endothermic . . .NH4NO3(s) + heat NH4+ (aq) + NO3- (aq)

4、Hsol = +25.7 kJ/molor have H = 0 . . . 4PROBABILITY - predictor of most stable stateWHY DO PROCESSES with H = 0 occur ?Consider expansion of gases to equal pressure:This is spontaneous because the final state,with equal # molecules in each flask, is much more probable than the initial state,with all

5、 molecules in flask 1, none in flask 2 SYSTEM CHANGES to state of HIGHER PROBABILITYFor entropy-driven reactions - the more RANDOM state.5Gas expansion - spontaneity from greater probabilityConsider distribution of 4 molecules in 2 flasksP1 P2P1 = P2With more molecules (1020) P1=P2 is most probable

6、by far6Directionality of ReactionsHow probable is it that reactant molecules will react? PROBABILITY suggests that a product-favored reaction will result in the dispersal of energy or dispersal of matter or both.7Spontaneous ProcessesA process that is spontaneous in one direction is not spontaneous

7、in the opposite direction.The direction of a spontaneous process can depend on temperature: Ice turning to water is spontaneous at T 0C, Water turning to ice is spontaneous at T S (liquids) S (solids)So (J/Kmol)H2O (g)188.8H2O (l) 69.9H2O (s) 47.9IceWaterVapourEntropy and Phase10The entropy of a sub

8、stance increases with temperature.Molecular motions different temps.Entropy and TemperatureHigher T means : more randomness larger S11Entropy and complexityIncrease in molecular complexity generally leads to increase in S.So (J/Kmol)CH4248.2C2H6336.1 C3H8419.412Ionic Solids : Entropy depends on exte

9、nt of motion of ions. This depends on the strength of coulombic attraction.Entropy of Ionic SubstancesEntropy increases when a pure liquid or solid dissolves in a solvent.NH4NO3(s) NH4+ (aq) + NO3- (aq)Ssol =ion pairsSo (J/Kmol)MgOMg2+ / O2-26.9NaFNa+ / F-51.5So(aq. ions) - So(s) = 259.8 - 151.1= 10

10、8.7 J K-1 mol-113Entropy Changes for Phase ChangesFor a phase change, DS = q/Twhere q = heat transferred in phase changeFor H2O (liq) - H2O(g)DH = q = +40,700 J/mol14The Molecular Interpretation of Entropy15Consider 2 H2(g) + O2(g) 2 H2O(l)DSo = 2 So (H2O) - 2 So (H2) + So (O2)DSo = 2 mol (69.9 J/Km

11、ol) - 2 mol (130.7 J/Kmol) + 1 mol (205.3 J/Kmol)DSo = -326.9 J/KNote that there is a decrease in S because 3 mol of gas give 2 mol of liquid.Calculating S for a ReactionDSo = S So (products) - S So (reactants)If S DECREASES, why is this a SPONTANEOUS REACTION? 16The Molecular Interpretation of Entr

12、opyEnergy is required to get a molecule to translate, vibrate or rotate.The more energy stored in translation, vibration and rotation, the greater the degrees of freedom and the higher the entropy.In a perfect crystal at 0 K there is no translation, rotation or vibration of molecules. Therefore, thi

13、s is a state of perfect order.Third Law of Thermodynamics: the entropy of a perfect crystal at 0 K is zero.Entropy changes dramatically at a phase change.17 E = q + wThe Laws of Thermodynamics0. Two bodies in thermal equilibrium are at same T1. Energy can never be created or destroyed.2. The total e

14、ntropy of the UNIVERSE ( = system plus surroundings) MUST INCREASE in every spontaneous process. STOTAL = Ssystem + Ssurroundings 03. The entropy (S) of a pure, perfectly crystalline compound at T = 0 K is ZERO. (no disorder)ST=0 = 0 (perfect xll)182nd Law of ThermodynamicsA reaction is spontaneous

15、(product-favored) if DS for the universe is positive.DSuniverse = DSsystem + DSsurroundingsDSuniverse 0 for product-favored processFirst, calc. entropy created by matter dispersal (DSsystem)Next, calc. entropy created by energy dispersal (DSsurround)19Consider 2 H2(g) + O2(g) - 2 H2O(l)DSo = 2 So (H

16、2O) - 2 So (H2) + So (O2)DSo = 2 mol (69.9 J/Kmol) - 2 mol (130.7 J/Kmol) + 1 mol (205.3 J/Kmol)DSo = -326.9 J/KNote that there is a decrease in S because 3 mol of gas give 2 mol of liquid.Calculating DS for a ReactionDSo = So (products) - So (reactants)202 H2(g) + O2(g) - 2 H2O(l)DSosystem = -326.9

17、 J/K 2nd Law of Thermodynamics212 H2(g) + O2(g) - 2 H2O(liq)DSosystem = -326.9 J/K 2nd Law of Thermodynamics222 H2(g) + O2(g) - 2 H2O(liq)DSosystem = -326.9 J/KCan calc. that DHorxn = DHosystem = -571.7 kJ 2nd Law of Thermodynamics232 H2(g) + O2(g) - 2 H2O(liq)DSosystem = -326.9 J/KCan calc. that DH

18、orxn = DHosystem = -571.7 kJ 2nd Law of Thermodynamics242 H2(g) + O2(g) - 2 H2O(liq)DSosystem = -326.9 J/KCan calc. that DHorxn = DHosystem = -571.7 kJDSosurroundings = +1917 J/K 2nd Law of Thermodynamics252 H2(g) + O2(g) - 2 H2O(l)DSosystem = -326.9 J/KDSosurroundings = +1917 J/KDSouniverse = +1590

19、. J/KThe entropy of the universe is increasing, so the reaction is product-favored. 2nd Law of Thermodynamics262 H2(g) + O2(g) - 2 H2O(liq)DSosystem = -326.9 J/KDSosurroundings = +1917 J/KDSouniverse = +1590. J/KThe entropy of the universe is increasing, so the reaction is product-favored. 2nd Law o

20、f Thermodynamics27Gibbs Free Energy, GDSuniv = DSsurr + DSsysGibbs Free Energy, GDSuniv = DSsurr + DSsysGibbs Free Energy, GDSuniv = DSsurr + DSsysMultiply through by -TGibbs Free Energy, GD Suniv = D Ssurr + D SsysMultiply through by -T-T D Suniv = D Hsys - T D SsysDSuniv = -DHsysT + DSsys Gibbs Fr

21、ee Energy, GD Suniv = D Ssurr + D SsysMultiply through by -T-T D Suniv = D Hsys - T D Ssys-T D Suniv = change in Gibbs free energy for the universe = D GsystemGibbs Free Energy, GDSuniv = DSsurr + DSsysMultiply through by -T-TDSuniv = DHsys - TDSsys-TDSuniv = change in Gibbs free energy for the univ

22、erse = DGuniv = DGsystemUnder standard conditions DGo = DHo - TDSoDSuniv = -DHsysT + DSsys Gibbs Free Energy, GDGo = DHo - T DSoGibbs free energy change = DGo = total energy change for system - energy lost in disordering the systemIf reaction is exothermic (DHo 0), then DGo 0), and entropy decreases

23、 (DSo 0 and reaction is reactant-favored.Gibbs Free Energy, G DGo = DHo - TDSo DHo DSo DGo Reactionexo(-)increase(+)-Prod-favoredendo(+)decrease(-)+React-favoredexo(-)decrease(-)?T dependentendo(+)increase(+)?T dependentMethods of calculating GTwo methods of calculating DGoGorxn = S Gfo (products) -

24、 S Gfo (reactants)Determine DHorxn and DSorxn and use Gibbs equation.b) Use tabulated values of free energies of formation, DGfo.DGo = DHo - TDSo36ExampleAt what T is the following reaction spontaneous?Br2(l) Br2(g) where DH = 30.91 kJ/mol, DS = 93.2 J/mol.KAns:DG = DH - TDS37Try 298 K just to see:D

25、G = DH - TDSDG = 30.91 kJ/mol - (298K)(93.2 J/mol.K)DG = (30.91 - 27.78) kJ/mol = 3.13 kJ/mol 0Not spontaneous at 298 KBr2(l) Br2(g) where DH = 30.91 kJ/mol, DS = 93.2 J/mol.K38Example (cont.)At what T then?DG = DH - TDST = DH/DST = (30.91 kJ/mol) /(93.2 J/mol.K)= 0T = 331.65 K = 58.5oC39Calculating

26、 DGIn our previous example, we needed to determine DHrxn and DSrxn to determine DGrxnNow, DG is a state function; therefore, we can use known DG to determine DGrxn using:40Standard DG of Formation: DGfLike DHf and S, DGf is defined as the “change in free energy that accompanies the formation of 1 mo

27、le of that substance for its constituent elements with all reactants and products in their standard state.”Like DHf, DGf = 0 for an element in its standard state:Example: DGf (O2(g) = 041ExampleDetermine the DGrxn for the following:C2H4(g) + H2O(l) C2H5OH(l)Tabulated DGf from tables like Appendix D:

28、 DGf(C2H5OH(l) = -175 kJ/mol DGf(C2H4(g) = 68 kJ/mol DGf(H2O (l) = -237 kJ/mol42Example (cont.)Using these values:C2H4(g) + H2O(l) C2H5OH(l)DGrxn = DGf(C2H5OH(l) DGf(C2H4(g) + DGf(H2O (l)DGrxn = -175 kJ 68 kJ + (-237 kJ)DGrxn = -6 kJ 0 ; therefore, spontaneous43More DG CalculationsSimilar to DH, one

29、 can use the DG for various reactions to determine DG for the reaction of interest (a “Hess Law” for DG)Example:C(s, diamond) + O2(g) CO2(g) DG = -397 kJC(s, graphite) + O2(g) CO2(g) DG = -394 kJ44More DG Calculations (cont.)C(s, diamond) + O2(g) CO2(g) DG = -397 kJC(s, graphite) + O2(g) CO2(g) DG =

30、 -394 kJCO2(g) C(s, graphite) + O2(g) DG = +394 kJC(s, diamond) C(s, graphite) DG = -3 kJDGrxn 0.rxn is spontaneous45DGrxn Reaction RateAlthough DGrxn can be used to predict if a reaction will be spontaneous as written, it does not tell us how fast a reaction will proceed.Example: C(s, diamond) + O2

31、(g) CO2(g)DGrxn = -397 kJBut diamonds are forever. 2 CO2(g) + H2O(g)Use enthalpies of formation to calculate DHorxn = -1238 kJ 0Use standard molar entropies to calculate DSorxn = -97.4 J/K = -0.0974 kJ/K 0 DGorxn = -1238 kJ - (298 K)(-0.0974 J/K) = -1209 kJ 0Reaction is product-favored in spite of n

32、egative DSorxn. Reaction is “enthalpy driven”Calculating DGGo for COUPLED CHEMICAL REACTIONSReduction of iron oxide by CO is an example ofusing TWO reactions coupled to each other in orderto drive a thermodynamically forbidden reaction:Fe2O3(s) 4 Fe(s) + 3/2 O2(g) DGorxn = +742 kJ 3/2 C(s) + 3/2 O2

33、(g) 3/2 CO2(g) DGorxn = -592 kJwith a thermodynamically allowed reaction:Overall :Fe2O3(s) + 3/2 C(s) 2 Fe(s) + 3/2 CO2(g)DGorxn= +301 kJ 25oCBUT DGorxn 563oC See Kotz, pp933-935 for analysis of the thermite reaction48Other examples of coupled reactions:Copper smelting Cu2S (s) 2 Cu (s) + S (s) DGor

34、xn= +86.2 kJ(FORBIDDEN) Couple this with:S (s) + O2 (g) SO2 (s) DGorxn= -300.1 kJ Overall: Cu2S (s) + O2 (g) 2 Cu (s) + SO2 (s) DGorxn= +86.2 kJ + -300.1 kJ = -213.9 kJ (ALLOWED)Coupled reactions VERY COMMON in Biochemistry :e.g.all bio-synthesis driven by ATP ADP for which DHorxn = -20 kJDSorxn = +

35、34 J/KDGorxn = -30 kJ 37oC49The Concentration Dependence of SpontaneityAs with H and S, G values have been tabulated for the standard statevery few reactions occur at standard conditionsnoting that H and S change very little with changing T, we can make the assumption50The Concentration Dependence o

36、f SpontaneityThis works for changes in T, but G will change significantly with changing concentration and/or pressureFor any reactionaA + bB cC + dD5152The Concentration Dependence of SpontaneityOne possible source of acid rain is the reaction between NO2, a pollutant from automobile exhausts, and w

37、ater.3 NO2(g) + H2O(l) 2 HNO3(g) + NO(g)Determine whether this is thermodynamically feasible (a) under standard conditions and (b) at 298 K, with each product gas present at P = 1.0010-6 atm. Given: Gf(NO2(g) = 51.3 kJ/mol Gf(H2O(l) = -237.1 kJ/mol Gf(HNO3(g) = -73.5 kJ/mol Gf(NO(g) = 87.6 kJ/mol535

38、4The Temperature Dependence of SpontaneityG = H - T S55BioenergeticsThe basic processes of life can be thought of as making order out of disorderThis seems to go against the second lawTo create order, systems must release heat to the surroundingsLiving systems use large amounts of energy to survivem

39、ost common energy sources are carbohydrates and fats56BioenergeticsThe reaction of glucose with oxygen is highly spontaneousC6H12O6 + 6 O2 6 CO2 + 6 H2O G= -2870 kJ/mol as is the oxidation of palmitic acidC15H31COOH + 23 O2 16 CO2 + 16 H2O G = -9790 kJ/mol57BioenergeticsThese reactions release too m

40、uch energy for a cell to handlesome of the energy must be storedstored by converting ADP to ATP :ADP + H3PO4 ATP + H2O G = 30.6 kJ/mol58BioenergeticsCells use energy stored in ATP to drive nonspontaneous reactionsATP + H2O ADP + H3PO4 G = -30.6 kJ/molATP conversion can be coupled to other reactionst

41、ransfers energy from one reaction to anotherAmino Acid synthesis59Bioenergeticsglutamic acid + NH3 glutamine + H2O G = 14 +kJ/molATP + H2O ADP + H3PO4 G = -30.6 kJ/molglutamic acid + NH3 + ATP glutamine + ADP + H3PO4G = -16.6 kJ/mol60BioenergeticsCells are not 100% efficient in storing energy as ATP1 glucose molecule produces 36 ATP moleculesC6H12O6 + 6 O2 6 CO2 + 6 H2O G = -2870 kJ/mol36 ADP + 3