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1、117Chemical Equilibrium 化學平衡2Chapter Goals1.Basic Concepts2.The Equilibrium Constant 平衡常數3.Variation of Kc with the Form of the Balanced Equation 4.The Reaction Quotient 反應商5.Uses of the Equilibrium Constant, Kc6.Disturbing a System at Equilibrium: Predictions7.The Haber Process: A Commercial Applic
2、ation of Equilibrium8.Disturbing a System at Equilibrium: Calculations9.Partial Pressures and the Equilibrium Constant10.Relationship between Kp and Kc11.Heterogeneous Equilibria12.Relationship between Gorxn and the Equilibrium Constant13.Evaluation of Equilibrium Constants at Different Temperatures
3、3Basic Concepts基本概念Chemical reactions that can occur in either direction are called reversible reaction可逆反應Reversible reactions do not go to completion Reactants are not completely converted to products. (反應物不會完全轉成產物)They can occur in either directionSymbolically, this is represented as:aA(g)+bB(g )
4、 cC(g)+dD(g)When A and B react to form C and D at the same rate at which C and D react to form A and B, the system is equilibrium (當A與B反應形成C和D的速率與C及D反應形成A和B的速率相同時稱之為平衡)4Basic Concepts 基本概念Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate.化學平衡是指在可逆反應中，正逆反應速
5、率相等，反應物和生成物各組分濃度不再改變的狀態。 A chemical equilibrium is a reversible reaction that the forward reaction rate is equal to the reverse reaction rate (化學平衡為可逆反應其正向反應與反向反應的速率相同)Chemical equilibria are dynamic equilibria (動態動態平衡平衡)Molecules are continually reacting, even though the overall composition of the
6、reaction mixture does not change5Basic ConceptsOne example of a dynamic equilibrium can be shown using radioactive 131I as a tracer in a saturated PbI2 solution. (利用放射線碘131當作追蹤劑, 看放射線碘存在何處) 1. place solid PbI2* in a saturated PbI2 solution PbI2(s)* Pb2+(aq)+2I-(aq) 2. Stir for a few minutes, then fi
7、lter the solution some of the radioactive iodine will go into solutionH2O將固體PbI2置於水中,攪拌數分後,再經過過濾一些放射線碘存於溶液中6Basic ConceptsGraphically, this is a representation of the rates for the forward and reverse reactions for this general reactionaA(g)+bB(g ) cC(g)+dD(g)Equilibrium isestablished 達成平衡狀態反應開始7Bas
8、ic ConceptsOne of the fundamental ideas of chemical equilibrium is that equilibrium can be established from either the forward or reverse direction2SO2(g)+ O2(g ) 2SO3(g)0.02M達成平衡8Basic Concepts2SO2(g)+ O2(g ) 2SO3(g)0.400mol開始莫耳數0.200mol0-0.056mol反應改變莫耳數-0.028mol+0.056mol0.344mol反應後莫耳數0.172mol0.056
9、mol2SO2(g)+ O2(g ) 2SO3(g)0開始莫耳數00.500mol+0.424mol 反應改變莫耳數+0.212mol-0.424mol0.424mol反應後莫耳數0.212mol0.076mol2: 1: 22: 1: 22: 1: 22: 1: 29Basic Concepts2SO2(g)+ O2(g ) 2SO3(g)0.400MInitial conc.0.200M0-0.056MChange due to rxn-0.028M+0.056M0.344MEquilibrium concn平衡濃度平衡濃度0.172M0.056M2SO2(g)+ O2(g ) 2SO3(
10、g)0Initial conc.00.500M+0.424MChange due to rxn+0.212M-0.424M0.424MEquilibrium concn0.212M0.076M2: 1: 22: 1: 22: 1: 22: 1: 2In 1.00 liter container反應均為氣體,在固定體積10The Equilibrium ConstantFor a simple one-step mechanism reversible reaction such as:The rates of the forward and reverse reactions can be r
11、epresented as:Forward rate (正反應速率): Ratef = kfAB Reverse rate (逆反應速率): Rater = krCDA(g)+B(g ) C(g)+D(g)11The Equilibrium ConstantWhen system is at equilibrium 當系統達成平衡 Ratef = Rater 正反應速率=逆反應速率which represents the forward rate kfAB = krCD which rearranges to kf CD kr AB =Because the ratio of two cons
12、tants is a constant we can define a new constant as follows :kf krkc=kc =CDAB12The Equilibrium ConstantSimilarly, for the general reaction:we can define a constant: 平衡常數 KcKc =CcDdAaBbaA(g)+bB(g ) cC(g)+dD(g)Products 產物Reactants 反應物This expression is valid for all reactions 13The Equilibrium Constan
13、tKc is the equilibrium constant平衡常數 .Kc is defined for a reversible reaction at a given temperature as the product of the equilibrium concentrations (in M) of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation, divided by the product of the equilibr
14、ium concentrations (in M) of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.各物種的體積莫耳濃度均為平衡時的濃度。Kc的數值等於方程式中各生成物濃度的係數次方相乘後，再除以各反應物濃度的係數次方。定溫時無論反應的初濃度如何改變，只要達到平衡時，其平衡常數均相等。此常數的大小僅與物種、溫度有關，而與濃度、壓力的大小無關。14The Equilibrium ConstantExample 17-1: Write
15、equilibrium constant expressions for the following reactions at 500oC. All reactants and products are gases at 500oC. Kc =PCl3Cl2PCl5PCl5 PCl3 + Cl2H2 + l2 2HIKc =HI2I2 I24NH3 + 5O2 4NO + 6H2 OKc =NO4H2O6NH34O25The Equilibrium Constant15Example 17-1: Calculation of Kc Some nitrogen and hydrogen are
16、placed in an empty 5.00-liter container at 500oC. When equilibrium is established, 3.01mol of N2, 2.10 mol of H2, and 0.565 mol of NH3 are present. Evaluate Kc for the following reaction at 500oC.N2(g) + 3H2(g) 2NH3(g) Kc N2H23NH32=N2: 3.01mol/5L = 0.602 MH2: 2.10mol/5L = 0.420 MNH3: 0.565mol/5L = 0
17、.113 M=(0.602)(0.420)3(0.113)2=0.28616The Equilibrium ConstantExample 17-2: One liter of equilibrium mixture from the following system at a high temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate Kc for
18、 the reaction.PCl5 PCl3 + Cl20.028MEquil s M0.172M0.086MKc =PCl3Cl2PCl5=(0.172)(0.086)(0.028)Kc=0.53One liter17The Equilibrium ConstantExample 17-3: The decomposition of PCl5 was studied at another temperature. One mole of PCl5 was introduced into an evacuated 1.00 liter container. The system was al
19、lowed to reach equilibrium at the new temperature. At equilibrium 0.60 mole of PCl3 was present in the container. Calculate the equilibrium constant at this temperature.PCl5 PCl3 + Cl21.00MInitial 00Kc =PCl3Cl2PCl5-0.60MChange 0.60M0.60M0.40MEquilibrium concn0.60M0.60M=(0.60)(0.60)(0.40)At another t
20、emperature=0.9018The Equilibrium ConstantExample 17-4: At a given temperature 0.80 mole of N2 and 0.90 mole of H2 were placed in an evacuated 1.00-liter container. At equilibrium 0.20 mole of NH3 was present. Calculate Kc for the reaction.N2 + 3H2 2NH3 0.80MInitial 0.90M0-0.10MChange -0.30M0.20M0.70
21、MEquilibrium concn0.60M0.20M=(0.20)2(0.70)(0.60)3=0.26Kc N2H23NH32=N2: 0.8mole/1Liter = 0.8M H2: 0.9mol/1Liter = 0.9MNH3: 0.2mol/1Liter = 0.2M The Equilibrium Constant19Example 17-2: Calculation of Kc We put 10.0 mol of N2O into a 2-L container at some temperature, where it decomposes according toAt
22、 equilibrium, 2.20 moles of N2O remain, Calculate the value of Kc for the reaction2N2O (g) 2N2(g) + O2(g) Kc N2O2N22O2=Initial N2O: 10.0mol/2L = 5.0 M=(1.1)2(3.9)2(1.95)=24.5equili N2O: 2.20mol/2L = 1.1 M2N2O (g) 2N2(g) + O2(g) 5.0MInitial 00-3.9MChange +3.9M+1.95M1.1MEquilibrium concn3.9M1.95M20Var
23、iation of Kc with the Form of the Balanced EquationThe value of Kc depends upon how the balanced equation is written.From example 17-2 we have this reaction:This reaction has a Kc=PCl3Cl2/PCl5=0.53PCl5 PCl3 + Cl221Variation of Kc with the Form of the Balanced EquationExample 17-5: Calculate the equi
24、librium constant for the reverse reaction by two methods, i.e, the equilibrium constant for this reaction.Equil. s 0.172M 0.086M 0.028 MThe concentrations are from Example 17-2.PCl3 + Cl2 PCl5Kc =PCl3Cl2PCl5=(0.172)(0.086)(0.028)Kc=1.922Variation of Kc with the Form of the Balanced EquationLarge equ
25、ilibriumLarge equilibrium constants constants indicate that indicate that most of the reactantsmost of the reactants are are converted to converted to products. (products. (大的平衡常數表示大部分的反應物轉大的平衡常數表示大部分的反應物轉成產物成產物) )Small equilibrium constantsSmall equilibrium constants indicate that indicate that onl
26、y only small amounts of products are formed.small amounts of products are formed. ( (小平衡常數表示僅少數產物生成小平衡常數表示僅少數產物生成) )Kc =PCl3Cl2PCl5=1.9=(0.172)(0.086)(0.028)Kc =Kc1Kc =Kc1or=0.531=1.9平衡狀態可由任一方向達成，其與反應物(A,B) ，及生成物(C,D)之初濃度有關-反應物之濃度大於平衡濃度反應由反應物向生成物方向而達平衡-生成物之初濃度大於平衡濃度反應由生成物向反應物而達平衡Kc定溫下為常數，其值僅隨溫度改變而改變
27、不同之平衡狀態，平衡濃度值 (A、B、C、D)可以不同，但其比值恆等於KcKc值大小無法決定達成平衡之移動方向-值大：平衡時，生成物較反應物多-值小：平衡時，反應物較生成物24The Reaction Quotient 反應商數The mass action expression質量作用表示法 or reaction quotient反應商數has the symbol Q. Q has the same form as Kc (Q即是Kc的另一表示形式)The major difference between Q and Kc is that the concentrations used
28、in Q are not necessarily equilibrium values. (Q並不一定是達成平衡的濃度)Q =CcDdAaBdFor this general reactionaA(g)+bB(g ) cC(g)+dD(g)Not necessarily equilibrium concentrations25The Reaction QuotientWhy do we need another “equilibrium constant” that does not use equilibrium concentrations?Q will help us predict h
29、ow the equilibrium will respond to an applied stress. Q值可用於預期反應受到外力影響時 的反應方向To make this prediction we compare Q with Kc.26The Reaction QuotientWhen:QKc Reverse reaction predominates until equilibrium is established (反應向左)僅有反應物僅有產物27The Reaction QuotientExample 17-6: The equilibrium constant for the
30、 following reaction is 49 at 450oC. If 0.22 mole of I2, 0.22 mole of H2, and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the system be at equilibrium? If not, what must occur to establish equilibrium? The concentrations given in the problem are not necessarily equilibrium
31、s. We can calculate QH2 + l2 2HI0.22M0.22M0.66MQ = HI2I2H2=(0.22)(0.22)(0.66)2=9.0Q=9.0 but Kc=49QKcForward reaction predominates until equilibrium is established (反應會持續往右進行,直至達到平衡)28Uses of the Equilibrium Constant, KcExample 17-7: The equilibrium constant, Kc, is 3.00 for the following reaction at
32、 a given temperature. If 1.00 mole of SO2 and 1.00 mole of NO2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium?SO2(g) + NO2(g) SO3(g) +NO(g)MM00Initial-x M-x M+x M+x MChange(0.5-x)M(0.5-x)MxMxMEquilibriumKc =
33、SO2NO2SO3NO=(0.5-x)(0.5-x)(x)(x)=3.0=(0.5-x)2(x)21.73=0.5-xx0.865-1.73x=x x=0.316M=SO3=NOSO2=NO2=0.5-0.316=0.184M29Uses of the Equilibrium Constant, KcExample 17-8: The equilibrium constant is 49 for the following reaction at 450oC. If 1.00 mole of HI is put into an evacuated 1.00-liter container an
34、d allowed to reach equilibrium, what will be the equilibrium concentration of each substance?H2(g) + I2(g) 2HI(g)001.0MInitial+x M+x M-2x MChangex Mx M1.0-2x MEquilibriumKc =H2I2HI2= (x)(x)(1.0-2x)2= 497.0=x1.0-2x7.0x=1.0-2x x=0.11M=H2=I2HI=1.0-(2x0.11)=0.78M30Disturbing a System at Equilibrium: Pre
35、dictionsLeChateliers Principle - If a change of conditions (stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce the stress in reaching a new state of equilibrium勒沙特原理：一平衡系統中，加一影響此反應平衡之因素時，反應會向抵銷此影響因素的方向進行Some possible stresses to a system at equili
36、brium are:1.Changes in concentration of reactants or products.2.Changes in pressure or volume (for gaseous reactions)3.Changes in temperature增加反應物濃度或移除生成物時，平衡往生成物方向移動氣相反應中，增加壓力或減少反應體積，平衡則往莫耳數減少之方向移動31Disturbing a System at Equlibrium: PredictionsFor convenience we may express the amount of a gas in
37、terms of its partial pressure rather than its concentration. 以分壓表示To derive this relationship, we must solve the ideal gas equation理想氣體方程式.PV=nRTP=(n/V)RT because (n/V) has the units mol/L (濃度)P=MRTThus at constant T the partial pressure of a gas is directly proportional to its concentration定溫下,一氣體的
38、分壓與其濃度成正比32Disturbing a System at Equlibrium: PredictionsChanges in Concentration of Reactants and/or Products 改變反應物或產物的濃度 Also true for changes in pressure for reactions involving gases.Look at the following system at equilibrium at 450oC.H2(g) + I2(g) 2HI(g)=49=H2I2HI2Kc If some H2 is added, QKc (
39、分母變小,分子不變)This favors the reverse reaction (反應往左進行)Equilibrium will shift to the left or reactant side33Disturbing a System at Equlibrium: PredictionsChanges in Volume (體積改變) (and pressure for reactions involving gases)Predict what will happen if the volume of this system at equilibrium is changed b
40、y changing the pressure at constant temperature:2NO2(g) N2O4(g)=NO22N2O4Kc If the volume is decreased, which increased the pressure (體積減少,壓力增大), (V , P , NO2 and N2O4 )Q= (2N2O4)/(2NO2)2 = (2/4)Kc= (1/2) KcQKcThis favors the reactants or the reverse reaction (反應向左)34Disturbing a System at Equlibrium
41、: Predictions3Changing the Reaction Temperature (改變溫度)Consider the following reaction at equilibrium:2SO2(g) + O2(g) 2SO3(g) Horxn=-198kJ/mol Is heat a reactant or product in this reaction?Heat is a product of this reaction! (放熱反應當作產物)Increasing the reaction temperature (增加溫度) stresses the productsT
42、his favors the reactant or reverse reaction (反應向左)Decreasing the reaction temperature stresses the reactantsThis favors the product or forward reaction (反應向右)2SO2(g) + O2(g) 2SO3(g) +198kJ/molDisturbing a System at Equlibrium: Predictions若為放熱反應,提高溫度反應向左若為吸熱反應,提高溫度反應向右A+B C+D+ heatA+B+ heat C+D36Dist
43、urbing a System at Equlibrium: PredictionsIntroduction of a Catalyst 加入催化劑Catalysts decrease the activation energy of both the forward and reverse reaction equally (催化劑會同時降低正反應及負反應的活化能)Catalysts do not affect the position of equilibrium. (因此催化劑不會改變平衡狀態)The concentrations of the products and reactant
44、s will be the same whether a catalyst is introduced or notEquilibrium will be established faster with a catalyst (加入催化劑可加速反應達成平衡)37Disturbing a System at Equlibrium: PredictionsExample 17-9: Given the reaction below at equilibrium in a closed container at 500oC. How would the equilibrium be influenc
45、ed by the following?N2(g) + 3H2(g) 2NH3(g) Horxn=-92kJ/molFactorsEffect on reaction procedurea. Increasing the reaction temperatureb. Decreasing the reaction temperaturec. Increasing the pressure by decreasing the volumed. Increasing the concentration of H2e. Decreasing the concentration of NH3f. In
46、troduction a platinum catalyst Left Right Right Right RightNo effectKc N2H23NH32=Disturbing a System at Equlibrium: PredictionsExample 17-10: How will an increase in pressure (caused by decreasing the volume) affect the equilibrium in each of the following reactions?FactorsEffect on equilibriuma. H2
47、(g) + I2(g) 2HI(g)b. 4NH3(g)+ 5O2(g) 4NO(g)+6H2O(g)c. PCl3(g) + Cl2(g) PCl5(g)d. 2H2(g) + O2(g) 2H2O(g)No effect Left Right Right假設壓力增加兩倍,則濃度增加兩倍Q (2)2HI2(2)H2x(2)I2=a. = KcQ (2)4NH34x(2)5O25(2)4NO 4x(2)6H2O6=b. = 2KcQ (2)PCl5(2)PCl3x(2)Cl2=c. = 0.5KcQ (2)2H2O2(2)2H22x(2)O2=d. = 0.5Kc39Disturbing a
48、System at Equlibrium: PredictionsExample 17-11: How will an increase in temperature increase in temperature affect each of the following reactions?FactorsEffect on equilibriuma. 2NO2 (g) 2N2O4(g) Horxn0b. H2(g)+ Cl2(g) 2HCl(g)+92kJc. H2(g) + l2(g) 2HI(g) Horxn=25kJ Left Left Right40The Haber Process
49、: A Practical Application of EquilibriumThe Haber process is used for the commercial production of ammonia哈柏製氨法:為商業化產氨的方式This is an enormous industrial process in the US and many other countries.Ammonia is the starting material for fertilizer production.Look at Example 17-9. What conditions did we p
50、redict would be most favorable for the production of ammonia?41The Haber Process: A Practical Application of EquilibriumN2(g) + 3H2(g) 2NH3(g) Horxn=-92kJ/molFe & metal oxideN2 is obtained from liquid air; H2 obtain from coal gasThis reactions is run at T=450oC and P of N2 =200 to 1000atm G0 which i
51、s favorable H0 also favorable S0 which is unfavorable G= H-T S G 0反應自然發生However the reaction kinetics are very slow at low THabers solution to this dilemma1. Increase T to increase rate, but yield is decreased (反應向左)2. Increase reaction pressure to right3. Use excess N2 to right4. Remove NH3 periodi
52、cally to rightThe reaction system never reaches equilibrium because NH3 is removed . This increase the reaction yield and helps with the kinetics (由於不斷的移除產物氨,所以無法達成平衡)42The Haber Process: A Practical Application of EquilibriumThis diagram illustrates the commercial system devised for the Haber proce
53、ss.43Disturbing a System at Equilibrium: CalculationsTo help with the calculations, we must determine the direction that the equilibrium will shift by comparing Q with Kc.Example 17-12: An equilibrium mixture from the following reaction was found to contain 0.20 mol/L of A, 0.30 mol/L of B, and 0.30
54、 mol/L of C. What is the value of Kc for this reaction?A(g) B(g) + C(g) 0.20M 0.3 M0.3MEquilibrium= (0.2)(0.3)(0.3)=BCAKc=0.4544Disturbing a System at Equilibrium: CalculationsIf the volume of the reaction vessel were suddenly doubled while the temperature remained constant, what would be the new eq
55、uilibrium concentrations?1. Calculate Q, after the volume has been doubledA(g) B(g) + C(g) 0.10M 0.15M 0.15M= (0.10)(0.15)(0.15)=BCAQ=0.22體積加倍,濃度均減半= (0.2)(0.3)(0.3)=BCAKc=0.45QKc45Disturbing a System at Equilibrium: CalculationsSince QKc the reaction will shift to the right to reestablish the equil
56、ibrium. (OKc, 反應向右以達成另一平衡)2. Use algebra to represent the new concentrationsA(g) B(g) + C(g) 0.1M0.15M0.15MInitial-x M+x M+x MChange(0.1-x) M(0.15+x) M(0.15+x) MEquilibrium= (0.1-x)(0.15+x)2=0.45=BCAKc0.045-0.45x=0.0225+0.30x+x2x2+0.75x-0.0225=0(a+b)2 =a2+2ab+b246Disturbing a System at Equilibrium:
57、Calculationsx2+0.75x-0.0225=0ax2+bx+c=0x= -b b2-4ac 2ax= -0.75 (0.075)2-4(1)(-0.0225) 2x1x= -0.75 0.081 2x= -0.78 and 0.03M Since 0xKc thus the equilibrium shifts to the left or reactant side48Disturbing a System at Equilibrium: CalculationsSet up the algebraic expressions to determine the equilibri
58、um concentrationsA(g) B(g) + C(g) 0.40M0.60M0.60MInitial+x M-x M-x MChange(0.4+x) M(0.60-x) M(0.60-x) MEquilibrium= (0.4+x)(0.60-x)2=0.45=BCAKc0.18+0.45x=0.36-1.2x+x2x2-1.65x+0.18=0x= 1.65 (-1.65)2-4(1)(0.18) 2x1x= 1.65 1.42 2x= 1.5 and 0.12M Since 0x0.60 x=0.12M A=0.40+x=0.52 MB=C=0.60-x=0.48 M 反應向
59、左49Disturbing a System at Equilibrium: CalculationsExample 17-14: A 2.00 liter vessel in which the following system is in equilibrium contains 1.20 moles of COCl2, 0.60 moles of CO and 0.20 mole of Cl2. Calculate the equilibrium constant. CO(g) + Cl2(g) COCl2(g)0.6/20.2/21.2/2Equilibrium= (0.30)(0.1
60、0)(0.6)=COCl2COCl2Kc=200.3M0.1M0.6MEquilibrium50Disturbing a System at Equilibrium: CalculationsAn additional 0.80 mole of Cl2 is added to the vessel at the same temperature. Calculate the molar concentrations of CO, Cl2, and COCl2 when the new equilibrium is established.CO(g) + Cl2(g) COCl2(g)0.3M0
61、.1M0.6MOrig. Equil.+0.4M(Stress) Add0.30 M0.50 M0.6 MNew InitialQKc 反應向右-x M-x M+x MChange(0.3-x) M(0.50-x)M(0.60+x) MEquilibrium= (0.3-x)(0.5-x)(0.6+x)=COCl2COCl2Kc=200.80 mole of Cl2 in 2-liter vessel 0.4M of Cl2 = (0.30)(0.50)(0.6)Qc=420x2-17x+2.4=0x= 17 (17)2-4(20)(2.3) 2x20X=0.67 and 0.18Since
62、0x 0 disorder increases (which favors spontaneity).亂度增加,有利於自發性的反應 S Sliquid Ssolid Suniverse = Ssystem + Ssurroundings 0 增加亂度Entropy increase (Ssysytem0), WhenTemperature increaseVolume increaseMixing of substanceIncreasing particle numberMolecular size and complexityIonic compounds with similar for
63、mulas but different charges68Example Without doing a calculation, predict whether the entropy change will be positive or negativea)C2H6(g) +7/2 O2(g) 3H2O(g) + 2 CO2(g)b)3C2H2(g) C6H6(l)c)C6H12O6(s) + 6 O2(g) 6 CO2(g) + 2 H2O(l)a) S00b) S00d) Hg(l), Hg(s), Hg(g)e) C2H6(g), CH4(g) , C3H8(g) f) CaS(s)
64、, CaO(s)d) Hg(l) Hg(s) Hg(g)e) CH4(g) C2H6(g) C3H8(g) f) CaO(s) 0 reaction is nonspontaneous (不自發性反應) G = 0 system is at equilibrium (系統達到平衡) G 0 reaction is spontaneous (自發性反應)76Free Energy Change, G, and SpontaneityChanges in free energy obey the same type of relationship we have described for ent
65、halpy, H, and entropy, S, changes. Grxn= n Gproducts - nGreactants00077Free Energy Change, G, and SpontaneityExample 15-17: Calculate Go298 for the reaction in Example 15-8. Use appendix K.C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) Grxn= 3 GfCO2(g) + 4 GfH2O (g) - GfC3H8 (g) + 5 GfO2 (g) 00000= 3(-394.4)+4(
66、-237.3) (-23.49)+5(0) kJ/mol= -2108.9 kJ/mol Go298 0, and the reaction is nonspontaneous at standard state conditions.78Relationship Between Gorxn and the Equilibrium Constant l Gorxn is the standard free energy change 標準狀態下的自由能變化u Gorxn is defined for the complete conversion of all reactants to all
67、 products. (被定義為所有反應物完全轉為產物)l Grxn is the free energy change at nonstandard conditions ( Grxn 則為非標準狀態下自由能的變化) For example, concentrations other than 1 M or pressures other than 1 atm.(例如濃度非1M或壓力非1大氣壓)l Grxn is related to Gorxn by the following relationship.Grxn=Gorxn + RT lnQ orG=Go + 2.303 RT logQR
68、= universal gas constant (8.314J/molK)T= absolute temperature (絕對溫度)Q= reaction quotient (反應商)79Relationship Between Gorxn and the Equilibrium Constant當達成平衡時, G=0 and Q=Kc. Then we can derive this relationship:0=Go + RT lnK or0=Go + 2.303 RT logKWhich rearranges to:Go = -RT lnK orGo = -2.303 RT log
69、K反應物及生成物均為氣體時, K為 Kp;反應物及生成物均為溶液時, K為 Kc;若為異相時,均可使用80Relationship Between Gorxn and the Equilibrium ConstantFor the following generalized reaction, the thermodynamic equilibrium constant is defined as follows:K =(aC)c(ad)d(aa)a(ab)baA(g)+bB(g ) cC(g)+dD(g)Whereaa is the activity of A ab is the activ
70、ity of B ac is the activity of C ad is the activity of D 81Relationship Between Gorxn and the Equilibrium ConstantThe relationships among Gorxn, K, and the spontaneity of a reaction are: GorxnKSpontaneity at unit concentration 1Forward reaction spontaneous= 0= 1System at equilibrium 0 1Reverse react
71、ion spontaneous82Relationship Between Gorxn and the Equilibrium Constant83Relationship Between Gorxn and the Equilibrium ConstantExample 17-17: Calculate the equilibrium constant, Kp, for the following reaction at 25oC from thermodynamic data in Appendix K.Note: this is a gas phase reaction.N2O4(g)
72、2NO2(g) 1. Calculate Gorxn Gorxn =2 GofNO2(g) - GofN2O4(g) Gorxn =2 (51.30kJ) (97.82kJ) Gorxn =4.78 kJ/mol rxn Gorxn =4.78x103 j/mol rxnThis reaction is nonspontaneous84Relationship Between Gorxn and the Equilibrium Constant2. Calculate K from Gorxn =-RT lnKplnKp = Gorxn/-RT =(4.78x103J/mol)/-(8.314
73、J/molK)(298K) =-1.93Kp = e-1.93 =0.145= (PNO2)2/(PN2O4)85Relationship Between Gorxn and the Equilibrium ConstantKp for the reverse reaction at 25oC can be calculated easily, it is the reciprocal of the above reaction.2NO2(g) N2O4(g) Gorxn =-4.78 kJ/molKp = 1/Kp =1/0.145 = 6.9 = (PN2O4)/(PNO2)286Relationship Between Gorxn and the Equilibrium ConstantExample 17-18: At 25oC and 1.00 atmosphere, Kp = 4.3 x 10-13 for the decomposition of NO2. Calculate Gorxn at 25oC.
