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1、Entropy and Free Energy熵与自由能,12、作为干燥介质的湿空气为什么要先经预热后再送入干燥器？教学重点：抓住重点词句理解课文内容, 深入了解阿联酋人民那样爱护花草树木的原因，体会阿联酋人民为了美化生活，付出的艰辛劳动和不懈努力。认识保护环境的重要性，激发学生热爱大自然的感情。,Entropy and Free Energy熵与自由能1Entropy and Free Energy熵与自由能12、作为干燥介质的湿空气为什么要先经预热后再送入干燥器？教学重点：抓住重点词句理解课文内容, 深入了解阿联酋人民那样爱护花草树木的原因，体会阿联酋人民为了美化生活，付出的艰辛劳动和不懈
2、努力。认识保护环境的重要性，激发学生热爱大自然的感情。Entropy and Free 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 spontane
3、ous; others never occur. WHY ?THERMODYNAMICSKINETICS2,1,Entropy and Free Energy熵与自由能12,Entropy-and-Free-Energy熵与自由能课件,Product-Favored Reactions,e.g. thermite reactionFe2O3(s) + 2 Al(s) 2 Fe(s) + Al2O3(s)DH = - 848 kJ,In general, product-favored reactions are exothermic.,3,Product-Favored Reactionse.
4、g.,Non-exothermic spontaneous reactions,But many spontaneous reactions or processes are endothermic . . .,NH4NO3(s) + heat NH4+ (aq) + NO3- (aq)Hsol = +25.7 kJ/mol,or have H = 0 . . .,4,Non-exothermic spontaneous rea,PROBABILITY - predictor of most stable state,WHY DO PROCESSES with H = 0 occur ?Con
5、sider 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 molecules in flask 1, none in flask 2,SYSTEM CHANGES to state of HIGHER PROBABILITYFor entropy-driven reactions - the mo
6、re RANDOM state.,5,PROBABILITY - predictor of mos,Gas expansion - spontaneity from greater probability,Consider distribution of 4 molecules in 2 flasks,P1 P2,P1 P2,P1 = P2,With more molecules (1020) P1=P2 is most probable by far,6,Gas expansion - spontaneity fr,Directionality of Reactions,How probab
7、le 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.,7,Directionality of ReactionsHow,Spontaneous Processes,A process that is spontaneous in one direction is not spontaneous in the opp
8、osite 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 0C.,8,Spontaneous ProcessesA process,Standard Entropies, So,Every substance at a given temperature and in a specific phase has a wel
9、l-defined Entropy At 298o the entropy of a substance is called So - with UNITS of J.K-1.mol-1 The larger the value of So, the greater the degree of disorder or randomness e.g. So (in J K-1 mol-1) : Br2 (liq) = 152.2 Br2 (gas) = 245.5For any process:,So = So(final) - So(initial),So(vap., Br2) = (245.
10、5-152.2) = 93.3 J K-1 mol-1,9,Standard Entropies, So Every s,S (gases) S (liquids) S (solids),So (J/Kmol)H2O (g)188.8H2O (l) 69.9H2O (s) 47.9,Entropy and Phase,10,S (gases) S (liquids) ,The entropy of a substance increases with temperature.,Molecular motions different temps.,Entropy and Temperature,
11、Higher T means : more randomness larger S,11,The entropy of a substance inc,Entropy and complexity,Increase in molecular complexity generally leads to increase in S.,So (J/Kmol)CH4248.2C2H6336.1 C3H8419.4,12,Entropy and complexityIncrease,Ionic Solids : Entropy depends on extent of motion of ions. T
12、his depends on the strength of coulombic attraction.,Entropy of Ionic Substances,Entropy 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-51.5,So(aq. ions) - So(s) = 259.8 - 151.1= 108.7 J K-1 mol-1,13
13、,Ionic Solids : Entropy depends,Entropy Changes for Phase Changes,For a phase change, DS = q/Twhere q = heat transferred in phase changeFor H2O (liq) - H2O(g)DH = q = +40,700 J/mol,14,Entropy Changes for Phase Chan,The Molecular Interpretation of Entropy,15,The Molecular Interpretation o,Consider 2
14、H2(g) + O2(g) 2 H2O(l)DSo = 2 So (H2O) - 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 S for a Reaction,DSo = S So (products) - S So (reactants),If S
15、DECREASES, why is this a SPONTANEOUS REACTION?,16,Consider 2 H2(g) + O2(g) 2 H,The Molecular Interpretation of Entropy,Energy 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
16、 the entropy.In a perfect crystal at 0 K there is no translation, rotation or vibration of molecules. Therefore, this 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,The Molecular Interpretati
17、on o, E = q + w,The Laws of Thermodynamics,0. Two bodies in thermal equilibrium are at same T1. Energy can never be created or destroyed.,2. The total entropy of the UNIVERSE ( = system plus surroundings) MUST INCREASE in every spontaneous process., STOTAL = Ssystem + Ssurroundings 0,3. The entropy
18、(S) of a pure, perfectly crystalline compound at T = 0 K is ZERO. (no disorder),ST=0 = 0 (perfect xll),18, E = q + wThe Laws of Thermo,2nd Law of Thermodynamics,A reaction is spontaneous (product-favored) if DS for the universe is positive.DSuniverse = DSsystem + DSsurroundingsDSuniverse 0 for produ
19、ct-favored processFirst, calc. entropy created by matter dispersal (DSsystem)Next, calc. entropy created by energy dispersal (DSsurround),19,2nd Law of ThermodynamicsA rea,Consider 2 H2(g) + O2(g) - 2 H2O(l)DSo = 2 So (H2O) - 2 So (H2) + So (O2)DSo = 2 mol (69.9 J/Kmol) - 2 mol (130.7 J/Kmol) + 1 mo
20、l (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 Reaction,DSo = So (products) - So (reactants),20,Consider 2 H2(g) + O2(g) -,2 H2(g) + O2(g) - 2 H2O(l)DSosystem = -326.9 J/K,2nd Law of Thermodynamics,21,2 H2(g) + O2(g)
21、 - 2 H2O(l)2,2 H2(g) + O2(g) - 2 H2O(liq)DSosystem = -326.9 J/K,2nd Law of Thermodynamics,22,2 H2(g) + O2(g) - 2 H2O(liq,2 H2(g) + O2(g) - 2 H2O(liq)DSosystem = -326.9 J/KCan calc. that DHorxn = DHosystem = -571.7 kJ,2nd Law of Thermodynamics,23,2 H2(g) + O2(g) - 2 H2O(liq,2 H2(g) + O2(g) - 2 H2O(li
22、q)DSosystem = -326.9 J/KCan calc. that DHorxn = DHosystem = -571.7 kJ,2nd Law of Thermodynamics,24,2 H2(g) + O2(g) - 2 H2O(liq,2 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 Thermodynamics,25,2 H2(g) + O2(g) - 2 H
23、2O(liq,2 H2(g) + O2(g) - 2 H2O(l)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 of Thermodynamics,26,2 H2(g) + O2(g) - 2 H2O(l)2,2 H2(g) + O2(g) - 2 H2O(liq)DSosystem = -326.9 J/KDSosurro
24、undings = +1917 J/KDSouniverse = +1590. J/KThe entropy of the universe is increasing, so the reaction is product-favored.,2nd Law of Thermodynamics,27,2 H2(g) + O2(g) - 2 H2O(liq,Gibbs Free Energy, G,DSuniv = DSsurr + DSsys,Gibbs Free Energy, GDSuniv = D,Gibbs Free Energy, G,DSuniv = DSsurr + DSsys,
25、Gibbs Free Energy, GDSuniv = D,Gibbs Free Energy, G,DSuniv = DSsurr + DSsysMultiply through by -T,Gibbs Free Energy, GDSuniv = D,Gibbs Free Energy, G,D Suniv = D Ssurr + D SsysMultiply through by -T-T D Suniv = D Hsys - T D Ssys,D,S,univ,=,-D,H,sys,T,+,D,S,sys,Gibbs Free Energy, GD Suniv =,Gibbs Fre
26、e Energy, G,D 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 Gsystem,Gibbs Free Energy, GD Suniv =,Gibbs Free Energy, G,DSuniv = DSsurr + DSsysMultiply through by -T-TDSuniv = DHsys - TDSsys-TDSuniv = change i
27、n Gibbs free energy for the universe = DGuniv = DGsystemUnder standard conditions DGo = DHo - TDSo,D,S,univ,=,-D,H,sys,T,+,D,S,sys,Gibbs Free Energy, GDSuniv = D,Gibbs Free Energy, G,DGo = DHo - T DSoGibbs free energy change = DGo = total energy change for system - energy lost in disordering the sys
28、temIf reaction is exothermic (DHo 0), then DGo 0), and entropy decreases (DSo 0 and reaction is reactant-favored.,Gibbs Free Energy, GDGo = DHo,Gibbs Free Energy, G,DGo = DHo - TDSo DHo DSo DGo Reactionexo(-)increase(+)-Prod-favoredendo(+)decrease(-)+React-favoredexo(-)decrease(-)?T dependentendo(+)
29、increase(+)?T dependent,Gibbs Free Energy, G DGo,Methods of calculating G,Two methods of calculating DGo,Gorxn = S Gfo (products) - S Gfo (reactants),Determine DHorxn and DSorxn and use Gibbs equation.b) Use tabulated values of free energies of formation, DGfo.,DGo = DHo - TDSo,36,Methods of calcula
30、ting GTwo m,Example,At what T is the following reaction spontaneous?,Br2(l) Br2(g),where DH = 30.91 kJ/mol, DS = 93.2 J/mol.K,Ans:,DG = DH - TDS,37,ExampleAt what T is the follow,Try 298 K just to see:,DG = DH - TDS,DG = 30.91 kJ/mol - (298K)(93.2 J/mol.K),DG = (30.91 - 27.78) kJ/mol = 3.13 kJ/mol,
31、0,Not spontaneous at 298 K,Br2(l) Br2(g) where DH = 30.91 kJ/mol, DS = 93.2 J/mol.K,38,Try 298 K just to see:DG = DH,Example (cont.),At what T then?,DG = DH - TDS,T = DH/DS,T = (30.91 kJ/mol) /(93.2 J/mol.K),= 0,T = 331.65 K = 58.5oC,39,Example (cont.)At what T then?,Calculating DG,In our previous e
32、xample, we needed to determine DHrxn and DSrxn to determine DGrxn,Now, DG is a state function; therefore, we can use known DG to determine DGrxn using:,40,Calculating DGIn our previous,Standard DG of Formation: DGf,Like DHf and S, DGf is defined as the “change in free energy that accompanies the for
33、mation of 1 mole 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) = 0,41,Standard DG of Formation: DGf,Example,Determine the DGrxn for the following:,C2H4(g) + H2O(l) C
34、2H5OH(l),Tabulated DGf from tables like Appendix D: DGf(C2H5OH(l) = -175 kJ/mol DGf(C2H4(g) = 68 kJ/mol DGf(H2O (l) = -237 kJ/mol,42,ExampleDetermine the DGrxn fo,Example (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 + (-
35、237 kJ),DGrxn = -6 kJ, 0 ; therefore, spontaneous,43,Example (cont.)Using these val,More DG Calculations,Similar to DH, one 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 kJ,C(s, graphite) + O2(g
36、) CO2(g) DG = -394 kJ,44,More DG CalculationsSimilar t,More DG Calculations (cont.),C(s, diamond) + O2(g) CO2(g) DG = -397 kJ,C(s, graphite) + O2(g) CO2(g) DG = -394 kJ,CO2(g) C(s, graphite) + O2(g) DG = +394 kJ,C(s, diamond) C(s, graphite) DG = -3 kJ,DGrxn 0.rxn is spontaneous,45,More DG Calculatio
37、ns (cont.)C,DGrxn Reaction Rate,Although 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(g) CO2(g),DGrxn = -397 kJ,But diamonds are forever.,0,DGrxn rate,46,DGrxn Reaction RateAlthough,Combus
38、tion of acetyleneC2H2(g) + 5/2 O2(g) - 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 negative DSor
39、xn. Reaction is “enthalpy driven”,Calculating DG,Combustion of acetyleneCalcula,Go for COUPLED CHEMICAL REACTIONS,Reduction 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 = +74
40、2 kJ,3/2 C(s) + 3/2 O2 (g) 3/2 CO2(g) DGorxn = -592 kJ,with a thermodynamically allowed reaction:,Overall :,Fe2O3(s) + 3/2 C(s) 2 Fe(s) + 3/2 CO2(g),DGorxn= +301 kJ 25oC,BUT DGorxn 563oC,See Kotz, pp933-935 for analysis of the thermite reaction,48,Go for COUPLED CHEMICAL REACT,Other examples of coup
41、led reactions:,Copper smelting,Cu2S (s) 2 Cu (s) + S (s) DGorxn= +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-
42、synthesis driven by ATP ADP for which DHorxn = -20 kJDSorxn = +34 J/KDGorxn = -30 kJ 37oC,49,Other examples of coupled reac,The Concentration Dependence of Spontaneity,As with H and S, G values have been tabulated for the standard statevery few reactions occur at standard conditionsnoting that H and
43、 S change very little with changing T, we can make the assumption,50,The Concentration Dependence,The Concentration Dependence of Spontaneity,This works for changes in T, but G will change significantly with changing concentration and/or pressureFor any reactionaA + bB cC + dD,51,The Concentration D
44、ependence,52,52,The Concentration Dependence of Spontaneity,One possible source of acid rain is the reaction between NO2, a pollutant from automobile exhausts, and water.3 NO2(g) + H2O(l) 2 HNO3(g) + NO(g)Determine whether this is thermodynamically feasible (a) under standard conditions and (b) at 2
45、98 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/mol,53,The Concentration Dependence,54,54,The Temperature Dependence of Spontaneity,G = H - T S,55,The Temperature Dependence of,Bioenergetic
46、s,The 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 survivemost common energy sources are carbohydrates and fats,56,Bioenerget
47、icsThe basic process,Bioenergetics,The 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/mol,57,BioenergeticsThe reaction of g,Bioenergetics,These reactions release too
48、 much 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/mol,58,BioenergeticsThese reactions r,Bioenergetics,Cells use energy stored in ATP to drive nonspontaneous reactionsATP + H2O ADP + H3PO4 G = -30.6 kJ/molATP conversio
49、n can be coupled to other reactionstransfers energy from one reaction to anotherAmino Acid synthesis,59,BioenergeticsCells use energy,Bioenergetics,glutamic 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/mo
50、l,60,Bioenergeticsglutamic acid + N,Bioenergetics,Cells 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 + 36 H3PO4 36 ATP + 36 H2O G=1102 kJ/molC6H12O6 + 6 O2 + 36 ADP + 36 H3PO4 6 CO2 + 6 H2O + 36 ATP + 3
