Chapter 17 Chemical equilibrium17章化学平衡

1、121.Basic Concepts2.The Equilibrium Constant 平衡常數平衡常數3.Variation of Kc with the Form of the Balanced Equation 4.The Reaction Quotient 反應商反應商5.Uses of the Equilibrium Constant, Kc6.Disturbing a System at Equilibrium: Predictions7.The Haber Process: A Commercial Application of Equilibrium8.Disturbing

2、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 Temperatures3Chemical reactions that can

3、occur in either direction are called reversible reaction可逆反應可逆反應 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反應物不會完全轉成產物) They can occur in either direction Symbolically, this is represented as:aA(g)+bB(g ) cC(g)+dD(g)When A and B reac

4、t to form C and D at the same rate at which C and D react to form A and B, the system is equilibrium (當當A與與B反應形成反應形成C和和D的速率與的速率與C及及D反應形成反應形成A和和B的速率相同時稱之為的速率相同時稱之為平衡平衡)4 Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate.化化學平衡學平衡是指在可逆反應中，正逆反應速率相等，反應物和是指在可逆反應

5、中，正逆反應速率相等，反應物和生成物各組分濃度不再改變的狀態。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 (化學平衡為化學平衡為可逆反應可逆反應其正向反其正向反應與反向反應的速率相同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 reac

6、ting, even though the overall composition of the reaction mixture does not change5 One example of a dynamic equilibrium can be shown using radioactive 131I as a tracer in a saturated PbI2 solution. (利用放射線碘利用放射線碘131當作追蹤劑當作追蹤劑, 看放射線碘存在何處看放射線碘存在何處) 1. place solid PbI2* in a saturated PbI2 solution PbI2

7、(s)* Pb2+(aq)+2I-(aq) 2. Stir for a few minutes, then filter the solution some of the radioactive iodine will go into solutionH2O將固體將固體PbI2置於水中置於水中,攪拌數分後攪拌數分後,再經過過濾再經過過濾一些放射線碘存於溶液中一些放射線碘存於溶液中6 Graphically, this is a representation of the rates for the forward and reverse reactions for this general r

8、eactionaA(g)+bB(g ) cC(g)+dD(g)Equilibrium isestablished 達成平衡狀態達成平衡狀態反應開始反應開始7 One 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達成平衡達成平衡82SO2(g)+ O2(g ) 2SO3(g)0.400mol開始莫耳數開始莫耳數0.200mo

9、l0-0.056mol反應改變莫耳數反應改變莫耳數-0.028mol+0.056mol0.344mol反應後莫耳數反應後莫耳數0.172mol0.056mol2SO2(g)+ O2(g ) 2SO3(g)0開始莫耳數開始莫耳數00.500mol+0.424mol 反應改變莫耳數反應改變莫耳數+0.212mol-0.424mol0.424mol反應後莫耳數反應後莫耳數0.212mol0.076mol92SO2(g)+ O2(g ) 2SO3(g)0.400MInitial conc.0.200M0-0.056MChange due to rxn-0.028M+0.056M0.344MEquili

10、brium concn平衡濃度平衡濃度0.172M0.056M2SO2(g)+ O2(g ) 2SO3(g)0Initial conc.00.500M+0.424MChange due to rxn+0.212M-0.424M0.424MEquilibrium concn0.212M0.076MIn 1.00 liter container反應均為氣體反應均為氣體,在固定體積在固定體積10 For a simple one-step mechanism reversible reaction such as: The rates of the forward and reverse react

11、ions can be represented as:Forward rate (正反應速率正反應速率): Ratef = kfAB Reverse rate (逆反應速率逆反應速率): Rater = krCDA(g)+B(g ) C(g)+D(g)11 When system is at equilibrium 當系統達成平衡當系統達成平衡 Ratef = Rater 正反應速率正反應速率=逆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

12、 ratio of two constants is a constant we can define a new constant as follows :kf krkc=kc =CDAB12 Similarly, for the general reaction:we can define a constant: 平衡常數平衡常數 KcKc =CcDdAaBbaA(g)+bB(g ) cC(g)+dD(g)Products 產物產物Reactants 反應物反應物This expression is valid for all reactions 13 Kc is the equilibr

13、ium constant平衡常數平衡常數 . 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 equilibrium concentrat

14、ions (in M) of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.各物種的各物種的體積莫耳濃度體積莫耳濃度均為平衡時的濃度。均為平衡時的濃度。Kc的數值等於的數值等於方程式中各方程式中各生成物濃度的係數次方相乘後，再除以各反應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物濃度的係數次方。定溫時定溫時無論反應的初濃度如何改變，無論反應的初濃度如何改變，只要達到平衡時，其只要達到平衡時，其平衡常數均相等平衡常數均相等。

15、 此常數的大小僅與此常數的大小僅與物種、溫度物種、溫度有關，而與有關，而與濃度、壓力濃度、壓力的大的大小無關。小無關。14 Example 17-1: Write 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 =NO4H2O6NH34O2515Examp

16、le 17-1: Calculation of Kc Some nitrogen and hydrogen are 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: 3.01mo

17、l/5L = 0.602 MH2: 2.10mol/5L = 0.420 MNH3: 0.565mol/5L = 0.113 M=(0.602)(0.420)3(0.113)2=0.28616Example 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 phosphor

18、us pentachloride. Calculate Kc for the reaction.PCl5 PCl3 + Cl20.028MEquil s M0.172M0.086MKc =PCl3Cl2PCl5=(0.172)(0.086)(0.028)Kc=0.53One liter17Example 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 sys

19、tem was allowed 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

20、 another temperature=0.9018Example 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 -0.30M0.20M0.70MEquilibrium c

21、oncn0.60M0.20M=(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 19Example 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 equilibrium, 2.20 moles of N2O remain

22、, 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 concn3.9M1.95M20 The value of Kc depends upon how the bal

23、anced equation is written. From example 17-2 we have this reaction: This reaction has a Kc=PCl3Cl2/PCl5=0.53PCl5 PCl3 + Cl221Example 17-5: Calculate the equilibrium 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 con

24、centrations are from Example 17-2.PCl3 + Cl2 PCl5Kc =PCl3Cl2PCl5=(0.172)(0.086)(0.028)Kc=1.922Kc =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，其與反應物平衡狀態可由任一方向達成，其與反應物(A,B) ，及生成物及生成物(C,D)之初濃度有關之初濃度有關-反應物之濃度大於平衡濃度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

25、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定溫下為常數，其值僅隨溫度改變而改變不同之平衡狀態，平衡濃度值不同之平衡狀態，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大：平衡時，生成物較反應物多-值小：平衡時，反應物較生成物值小：平衡時，反應物較生成物24 The mass action expression質量作用表示法質量作用表示法 or reaction quotient反

26、應商數反應商數has the symbol Q. Q has the same form as Kc (Q即是即是Kc的另一表示形式的另一表示形式) The major difference between Q and Kc is that the concentrations used in Q are not necessarily equilibrium values. (Q並不一定是達成平衡的濃度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 equilib

27、rium concentrations25 Why do we need another “equilibrium constant” that does not use equilibrium concentrations? Q will help us predict how the equilibrium will respond to an applied stress. Q值可用於預期反應受到值可用於預期反應受到外力影響時外力影響時 的反應方向的反應方向 To make this prediction we compare Q with Kc.26When:QKc Reverse r

28、eaction predominates until equilibrium is established (反應向左反應向左)僅有反應物僅有反應物僅有產物僅有產物27Example 17-6: The equilibrium constant for the 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 equil

29、ibrium? If not, what must occur to establish equilibrium? The concentrations given in the problem are not necessarily equilibrium 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

30、右進行反應會持續往右進行, ,直至達到平衡直至達到平衡)28Example 17-7: The equilibrium constant, Kc, is 3.00 for the following reaction at 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 a

31、t 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 =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=0.184M29Example 17-8: The equilibrium constant is 49 for the following reaction at

32、 450oC. If 1.00 mole of HI is put into an evacuated 1.00-liter container and 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=x1.0-2x7.0 x=1.0-2x

33、 x=0.11M=H2=I2HI=1.0-(2x0.11)=0.78M30LeChateliers 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，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

34、抵銷此影響因素的方向進行反應會向抵銷此影響因素的方向進行Some possible stresses to a system at equilibrium 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氣相反應中

35、，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31 For convenience we may express the amount of a gas in terms of its partial pressure rather than its concentration. 以分壓表示以分壓表示 To derive this relationship, we must solve the ideal gas equation理想氣體方程式理想氣體方程式.PV=nRTP=(n/V)RT because (n/V) has the units mol/L (濃度濃度)P=MRTThu

36、s at constant T the partial pressure of a gas is directly proportional to its concentration定溫下定溫下,一氣體的分壓與其濃度成正比一氣體的分壓與其濃度成正比32Changes in Concentration of Reactants and/or Products 改變反應物或產物的濃度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

37、equilibrium at 450oC.H2(g) + I2(g) 2HI(g)=49=H2I2HI2Kc If some H2 is added, QKc (分母變小分母變小,分子不變分子不變)This favors the reverse reaction (反應往左進行反應往左進行)Equilibrium will shift to the left or reactant side33Changes in Volume (體積改變體積改變) (and pressure for reactions involving gases)Predict what will happen if

38、the volume of this system at equilibrium is changed by changing the pressure at constant temperature:2NO2(g) N2O4(g)=NO22N2O4Kc If the volume is decreased, which increased the pressure (體積減少體積減少,壓力增大壓力增大), (V , P , NO2 and N2O4 )Q= (2N2O4)/(2NO2)2 = (2/4)Kc= (1/2) KcQKcThis favors the reactants or t

39、he reverse reaction (反應向左反應向左)343 Changing the Reaction Temperature (改變溫度改變溫度) 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! (放熱反應當作產物放熱反應當作產物)Increasing the reaction temperat

40、ure (增加溫度增加溫度) stresses the productsThis favors the reactant or reverse reaction (反應向左反應向左)Decreasing the reaction temperature stresses the reactantsThis favors the product or forward reaction (反應向右反應向右)2SO2(g) + O2(g) 2SO3(g) +198kJ/mol若為放熱反應若為放熱反應,提高溫度反應向左提高溫度反應向左若為吸熱反應若為吸熱反應,提高溫度反應向右提高溫度反應向右A+BC+

41、D+ heatA+B+ heat C+D36Introduction of a Catalyst 加入加入催化劑催化劑Catalysts decrease the activation energy of both the forward and reverse reaction equally (催化劑會同時降低正催化劑會同時降低正反應及負反應的活化能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因此催化劑不會改變平衡狀態)The concentrations of the prod

42、ucts and reactants will be the same whether a catalyst is introduced or notEquilibrium will be established faster with a catalyst (加加入催化劑可加速反應達成平衡入催化劑可加速反應達成平衡)37Example 17-9: Given the reaction below at equilibrium in a closed container at 500oC. How would the equilibrium be influenced by the follo

43、wing?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. Introduction a pl

44、atinum catalyst Left Right Right Right RightNo effectKc N2H23NH32=Example 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(g) + I2(g) 2HI(g)b. 4NH3(g)+ 5O2(g) 4NO(g)+6H2O(g)c. PCl3(g)

45、 + Cl2(g) PCl5(g)d. 2H2(g) + O2(g) 2H2O(g)No effect Left Right Right假設壓力增加兩倍假設壓力增加兩倍,則濃度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. = 0.5Kc39Example 17-11: How will an affect each of the following rea

46、ctions?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 Right40 The Haber process is used for the commercial production of ammonia哈柏製氨法哈柏製氨法:為商業化產氨的方式為商業化產氨的方式 This is an enormous industrial process in the US and many ot

47、her countries. Ammonia is the starting material for fertilizer production. Look at Example 17-9. What conditions did we predict would be most favorable for the production of ammonia?41N2(g) + 3H2(g) 2NH3(g) Horxn=-92kJ/molFe & metal oxideN2 is obtained from liquid air; H2 obtain from coal gasThi

48、s reactions is run at T=450oC and P of N2 =200 to 1000atm G0 which is favorable H0 also favorable S0 which is unfavorable G= H-T S G 0反應反應自自然發然發生生However the reaction kinetics are very slow at low THabers solution to this dilemma1. Increase T to increase rate, but yield is decreased (反應向左反應向左)2. Inc

49、rease reaction pressure to right3. Use excess N2 to right4. Remove NH3 periodically 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由於不斷的移除產物氨,所以無法達成平衡所以無法達成平衡)42This diagram illustrates the commer

50、cial system devised for the Haber process.43To 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 mol/L of

51、 C. What is the value of Kc for this reaction?A(g) B(g) + C(g) 0.20M 0.3 M 0.3MEquilibrium= (0.2)(0.3)(0.3)=BCAKc=0.4544 If the volume of the reaction vessel were suddenly doubled while the temperature remained constant, what would be the new equilibrium concentrations?1. Calculate Q, after the volu

52、me has been doubledA(g) B(g) + C(g) 0.10M 0.15M 0.15M= (0.10)(0.15)(0.15)=BCAQ=0.22體積加倍體積加倍,濃度均減半濃度均減半= (0.2)(0.3)(0.3)=BCAKc=0.45QKc45 Since QKc the reaction will shift to the right to reestablish the equilibrium. (OKc, 反應向右以達成另一平衡反應向右以達成另一平衡)2. Use algebra to represent the new concentrationsA(g) B

53、(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+0.30 x+x2x2+0.75x-0.0225=0(a+b)2 =a2+2ab+b246x2+0.75x-0.0225=0ax2+bx+c=0 x= -b b2-4ac 2ax= -0.75 (0.075)2-4(1)(-0.0225) 2x1x= -0.75 0.081 2x= -0.78 and 0.03M Sinc

54、e 0 xKc thus the equilibrium shifts to the left or reactant side48Set up the algebraic expressions to determine the equilibrium 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

55、+0.18=0 x= 1.65 (-1.65)2-4(1)(0.18) 2x1x= 1.65 1.42 2x= 1.5 and 0.12M Since 0 x0.60 x=0.12M A=0.40+x=0.52 MB=C=0.60-x=0.48 M 反應向左反應向左49Example 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

56、 equilibrium constant. CO(g) + Cl2(g) COCl2(g)0.6/20.2/21.2/2Equilibrium= (0.30)(0.10)(0.6)=COCl2COCl2Kc=200.3M0.1M0.6MEquilibrium50An 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 esta

57、blished.CO(g) + Cl2(g) COCl2(g)0.3M0.1M0.6MOrig. Equil.+0.4M(Stress) Add0.30 M0.50 M0.6 MNew InitialQKc 反應向右反應向右-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=420 x2-17x+2.4=0 x= 17

58、 (17)2-4(20)(2.3) 2x20X=0.67 and 0.18Since 0 x 0 disorder increases (which favors spontaneity).亂度增加亂度增加,有利於自發性的反應有利於自發性的反應 S Sliquid Ssolid Suniverse = Ssystem + Ssurroundings 0 增加亂度增加亂度 Entropy increase (Ssysytem0), When Temperature increase Volume increase Mixing of substance Increasing particle n

59、umber Molecular size and complexity Ionic compounds with similar formulas 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

60、) S00b) S00d) Hg(l), Hg(s), Hg(g)e) C2H6(g), CH4(g) , C3H8(g) f) CaS(s), CaO(s)d) Hg(l) Hg(s) Hg(g)e) CH4(g) C2H6(g) C3H8(g) f) CaO(s) 0 reaction is nonspontaneous (不自發性反應不自發性反應) G = 0 system is at equilibrium (系統達到平衡系統達到平衡) G 0 reaction is spontaneous (自發性反應自發性反應)76 Changes in free energy obey the sam