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Chemical Equilibrium 化學平衡

17. Chemical Equilibrium 化學平衡. Chapter Goals. Basic Concepts The Equilibrium Constant 平衡常數 Variation of K c with the Form of the Balanced Equation The Reaction Quotient 反應商 Uses of the Equilibrium Constant, K c Disturbing a System at Equilibrium: Predictions

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Chemical Equilibrium 化學平衡

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  1. 17 Chemical Equilibrium 化學平衡

  2. Chapter Goals • Basic Concepts • The Equilibrium Constant 平衡常數 • Variation of Kc with the Form of the Balanced Equation • The Reaction Quotient 反應商 • Uses of the Equilibrium Constant, Kc • Disturbing a System at Equilibrium: Predictions • The Haber Process: A Commercial Application of Equilibrium • Disturbing a System at Equilibrium: Calculations • Partial Pressures and the Equilibrium Constant • Relationship between Kp and Kc • Heterogeneous Equilibria • Relationship between Gorxn and the Equilibrium Constant • Evaluation of Equilibrium Constants at Different Temperatures

  3. Basic 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 direction • Symbolically, this is represented as: aA(g)+bB(g )  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)

  4. Basic 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(動態平衡) • Molecules are continually reacting, even though the overall composition of the reaction mixture does not change

  5. Basic Concepts • 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 filter the solution some of the radioactive iodine will go into solution H2O 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

  6. Basic Concepts • Graphically, this is a representation of the rates for the forward and reverse reactions for this general reaction aA(g)+bB(g )  cC(g)+dD(g) 反應開始 Equilibrium is established 達成平衡狀態

  7. Basic Concepts • One of the fundamental ideas of chemical equilibrium is that equilibrium can be established from either the forward or reverse direction 2SO2(g)+ O2(g )  2SO3(g) 達成平衡 0.02M

  8. Basic Concepts 開始莫耳數 0.400mol 0.200mol 0 2SO2(g)+ O2(g )  2SO3(g) -0.056mol -0.028mol +0.056mol 反應改變莫耳數 0.344mol 0.172mol 0.056mol 反應後莫耳數 2: 1: 2 開始莫耳數 0 0 0.500mol 2SO2(g)+ O2(g )  2SO3(g) +0.424mol +0.212mol -0.424mol 反應改變莫耳數 0.424mol 0.212mol 0.076mol 反應後莫耳數 2: 1: 2

  9. Basic Concepts 反應均為氣體,在固定體積 In 1.00 liter container Initial conc. 0.400M 0.200M 0 2SO2(g)+ O2(g )  2SO3(g) -0.056M -0.028M +0.056M Change due to rxn 0.344M 0.172M 0.056M Equilibrium conc’n 平衡濃度 2: 1: 2 Initial conc. 0 0 0.500M 2SO2(g)+ O2(g )  2SO3(g) +0.424M +0.212M -0.424M Change due to rxn 0.424M 0.212M 0.076M Equilibrium conc’n 2: 1: 2

  10. For a simple one-step mechanism reversible reaction such as: The rates of the forward and reverse reactions can be represented as: Forward rate(正反應速率):Ratef = kf[A][B] Reverse rate(逆反應速率):Rater = kr[C][D] The Equilibrium Constant A(g)+B(g )  C(g)+D(g)

  11. The Equilibrium Constant • When 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 kf[A][B] = kr[C][D] which rearranges to kf[C][D] kr [A][B] • Because the ratio of two constants is a constant we can define a new constant as follows : = = = [C][D] kf kc kc [A][B] kr

  12. Similarly, for the general reaction: we can define a constant: 平衡常數 Kc The Equilibrium Constant aA(g)+bB(g )  cC(g)+dD(g) Products 產物 Reactants 反應物 = • This expression is valid for all reactions [C]c[D]d Kc [A]a[B]b

  13. The Equilibrium Constant • Kc 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 equilibrium 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。

  14. The Equilibrium Constant • Example 17-1: Write equilibrium constant expressions for the following reactions at 500oC. All reactants and products are gases at 500oC. PCl5  PCl3 + Cl2 H2 + l2  2HI [HI]2 Kc [I2] [I2] = = = 4NH3 + 5O2  4NO + 6H2 O [PCl3][Cl2] Kc [NO]4[H2O]6 [PCl5] Kc [NH3]4[O2]5

  15. [NH3]2 Kc [N2][H2]3 The Equilibrium Constant Example 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 NH3are present. Evaluate Kc for the following reaction at 500oC. N2(g) + 3H2(g) 2NH3(g) [N2]: 3.01mol/5L = 0.602 M [H2]: 2.10mol/5L = 0.420 M = [NH3]: 0.565mol/5L = 0.113 M (0.113)2 0.286 = = (0.602)(0.420)3

  16. Example 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 the reaction. The Equilibrium Constant Equil []’s M 0.028M 0.172M 0.086M PCl5  PCl3 + Cl2 One liter = = (0.172)(0.086) [PCl3][Cl2] Kc Kc= 0.53 [PCl5] (0.028)

  17. The Equilibrium Constant Example 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 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. Initial 1.00M 0 0 0.60M 0.40M 0.60M Equilibrium conc’n PCl5  PCl3 + Cl2 Change -0.60M 0.60M 0.60M = = [PCl3][Cl2] (0.60)(0.60) K’c =0.90 [PCl5] (0.40) At another temperature

  18. Initial 0.80M 0.90M 0 0.20M 0.70M 0.60M Equilibrium conc’n [NH3]2 Kc [N2][H2]3 (0.20)2 (0.70)(0.60)3 The Equilibrium Constant Example 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: 0.8mole/1Liter = 0.8M H2: 0.9mol/1Liter = 0.9M NH3: 0.2mol/1Liter = 0.2M N2 + 3H2 2NH3 Change -0.10M -0.30M 0.20M = = =0.26

  19. Initial 5.0M 0 0 [N2]2[O2] Kc [N2O]2 The Equilibrium Constant Example 17-2: Calculation of Kc We put 10.0 mol of N2O into a 2-L container at some temperature, where it decomposes according to At equilibrium, 2.20 moles of N2O remain, Calculate the value of Kc for the reaction 2N2O (g)  2N2(g) + O2(g) Initial [N2O]: 10.0mol/2L = 5.0 M equili [N2O]: 2.20mol/2L = 1.1 M 2N2O (g)  2N2(g) + O2(g) = +1.95M Change -3.9M +3.9M 1.95M 1.1M 3.9M Equilibrium conc’n (3.9)2(1.95) 24.5 = = (1.1)2

  20. The value of Kc depends upon how the balanced equation is written. From example 17-2 we have this reaction: This reaction has a Kc=[PCl3][Cl2]/[PCl5]=0.53 Variation of Kc with the Form of the Balanced Equation PCl5  PCl3 + Cl2

  21. Example 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 0.028M The concentrations are from Example 17-2. [PCl5] K’c [PCl3][Cl2] (0.028) (0.172)(0.086) Variation of Kc with the Form of the Balanced Equation PCl3 + Cl2 PCl5 = = K’c= 1.9

  22. 1 1 1 Kc K’c = = = K’c Kc 0.53 [PCl5] K’c [PCl3][Cl2] (0.028) (0.172)(0.086) Variation of Kc with the Form of the Balanced Equation =1.9 or =1.9 • Large equilibriumconstants indicate that most of the reactants are converted to products. (大的平衡常數表示大部分的反應物轉成產物) • Small equilibrium constants indicate that only small amounts of products are formed.(小平衡常數表示僅少數產物生成) = =

  23. 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 • 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Kc • Kc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

  24. The 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 in Q are not necessarily equilibrium values.(Q並不一定是達成平衡的濃度) For this general reaction aA(g)+bB(g )  cC(g)+dD(g) = [C]c[D]d Not necessarily equilibrium concentrations Q [A]a[B]d

  25. The Reaction Quotient • 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外力影響時 的反應方向 • To make this prediction we compare Q with Kc.

  26. The Reaction Quotient 僅有產物 僅有反應物 When: Q<KcForward reaction predominates until equilibrium is established(反應向右) Q=KcThe system is at equilibrium (達成平衡) Q>Kc Reverse reaction predominates until equilibrium is established(反應向左)

  27. 0.22M 0.22M 0.66M (0.66)2 = (0.22)(0.22) The Reaction Quotient Example 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 equilibrium? If not, what must occur to establish equilibrium? The concentrations given in the problem are not necessarily equilibrium []’s. We can calculate Q H2 + l2 2HI = =9.0 [HI]2 Q=9.0 but Kc=49 Q<Kc Q [I2][H2] Forward reaction predominates until equilibrium is established(反應會持續往右進行,直至達到平衡)

  28. ½M (0.5-x)M ½M (0.5-x)M xM 0 xM 0 Equilibrium Initial Change -x M -x M +x M +x M [SO3][NO] (x)(x) (x)2 Kc = = [SO2][NO2] (0.5-x)(0.5-x) (0.5-x)2 x 0.5-x Uses of the Equilibrium Constant, Kc Example 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 at equilibrium? SO2(g)+ NO2(g) SO3(g)+NO(g) = =3.0 • 0.865-1.73x=x • x=0.316M=[SO3]=[NO] • [SO2]=[NO2]=0.5-0.316=0.184M 1.73=

  29. 0 0 1.0M Initial +x M Change +x M -2x M x M x M 1.0-2x M Equilibrium (1.0-2x)2 [HI]2 = = (x)(x) [H2][I2] 1.0-2x 7.0= x Uses of the Equilibrium Constant, Kc Example 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 and allowed to reach equilibrium, what will be the equilibrium concentration of each substance? H2(g)+ I2(g) 2HI(g) Kc =49 • 7.0x=1.0-2x • x=0.11M=[H2]=[I2] • [HI]=1.0-(2x0.11)=0.78M

  30. Disturbing a System at Equilibrium: Predictions • LeChatelier’s 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 equilibrium are: • Changes in concentration of reactants or products. • Changes in pressure or volume (for gaseous reactions) • 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

  31. Disturbing a System at Equlibrium: Predictions • 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 • To derive this relationship, we must solve the ideal gas equation理想氣體方程式. PV=nRT P=(n/V)RT because (n/V) has the units mol/L(濃度) P=MRT Thus at constant T the partial pressure of a gas is directly proportional to its concentration 定溫下,一氣體的分壓與其濃度成正比

  32. =49 [HI]2 Kc = [H2][I2] Disturbing a System at Equlibrium: Predictions Changes inConcentration 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) If some H2 is added, Q<Kc(分母變大,分子不變) This favors the forward reaction(反應往右移動) Equilibrium will shift to the right or product side If we remove some H2, Q>Kc(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 side

  33. [N2O4] Kc = [NO2]2 Disturbing a System at Equlibrium: Predictions Changes 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 by changing the pressure at constant temperature: 2NO2(g) N2O4(g) If the volume is decreased, which increased the pressure(體積減少,壓力增大), (V, P, [NO2] and [N2O4]) Q= (2[N2O4])/(2[NO2])2 = (2/4)Kc= (1/2) Kc Q<Kc This favors product formation or the forward reaction (反應向右) If the volume is increased, which decreased the pressure, Q>Kc This favors the reactants or the reverse reaction(反應向左)

  34. Disturbing a System at Equlibrium: Predictions • Changing 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 products This favors the reactant or reverse reaction(反應向左) Decreasing the reaction temperature stresses the reactants This favors the product or forward reaction(反應向右) 2SO2(g)+ O2(g) 2SO3(g) +198kJ/mol

  35. Disturbing a System at Equlibrium: Predictions 若為放熱反應,提高溫度反應向左 A+BC+D+ heat 若為吸熱反應,提高溫度反應向右 A+B+ heat C+D

  36. Introduction of a Catalyst 加入催化劑 Catalysts decrease the activation energy of both the forward and reverse reactionequally(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 reactants will be the same whether a catalyst is introduced or not Equilibrium will be established faster with a catalyst(加入催化劑可加速反應達成平衡) Disturbing a System at Equlibrium: Predictions

  37. [NH3]2 Kc [N2][H2]3 Disturbing a System at Equlibrium: Predictions Example 17-9: Given the reaction below at equilibrium in a closed container at 500oC. How would the equilibrium be influenced by the following? N2(g)+ 3H2(g) 2NH3(g)Horxn=-92kJ/mol Effect on reaction procedure Factors a. Increasing the reaction temperature b. Decreasing the reaction temperature c. Increasing the pressure by decreasing the volume d. Increasing the concentration of H2 e. Decreasing the concentration of NH3 f. Introduction a platinum catalyst  Left  Right  Right  Right  Right No effect =

  38. Disturbing a System at Equlibrium: Predictions Example 17-10: How will an increase in pressure (caused by decreasing the volume) affect the equilibrium in each of the following reactions? Effect on equilibrium Factors a. H2(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 a. b. (2)4[NO] 4x(2)6[H2O]6 (2)2[HI]2 Q Q = Kc = = 2Kc = (2)[H2]x(2)[I2] (2)4[NH3]4x(2)5[O2]5 c. d. (2)[PCl5] (2)2[H2O]2 Q = 0.5Kc = Q = 0.5Kc = (2)[PCl3]x(2)[Cl2] (2)2[H2]2x(2)[O2]

  39. Disturbing a System at Equlibrium: Predictions Example 17-11: How will an increase in temperature affect each of the following reactions? Effect on equilibrium Factors a. 2NO2 (g) 2N2O4(g) Horxn<0 b. H2(g)+ Cl2(g) 2HCl(g)+92kJ c. H2(g) + l2(g)  2HI(g) Horxn=25kJ  Left  Left  Right 39

  40. The Haber Process: A Practical Application of Equilibrium • The 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 predict would be most favorable for the production of ammonia?

  41. Fe & metal oxide N2(g)+ 3H2(g)2NH3(g)Horxn=-92kJ/mol The Haber Process: A Practical Application of Equilibrium • N2 is obtained from liquid air; H2 obtain from coal gas • This reactions is run at T=450oC and P of N2 =200 to1000atm • G<0 which is favorable H<0 also favorable • S<0 which is unfavorableG=H-TS △G < 0反應自然發生 • However the reaction kinetics are very slow at low T • Haber’s solution to this dilemma • 1. Increase T to increase rate, but yield is decreased(反應向左) • 2. Increase reaction pressure to  right • 3. Use excess N2 to  right • 4. Remove NH3 periodically to  right • The 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)

  42. The Haber Process: A Practical Application of Equilibrium This diagram illustrates the commercial system devised for the Haber process.

  43. Equilibrium 0.20M 0.3M 0.3M [B][C] Kc = [A] Disturbing a System at Equilibrium: Calculations To 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 C. What is the value of Kc forthis reaction? A(g) B(g) + C(g) (0.3)(0.3) = =0.45 (0.2)

  44. 0.10M 0.15M 0.15M [B][C] [B][C] Kc = Q = [A] [A] Disturbing a System at Equilibrium: Calculations • 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 volume has been doubled 體積加倍,濃度均減半 A(g) B(g) + C(g) (0.15)(0.15) =0.22 = (0.10) Q<Kc (0.3)(0.3) = =0.45 (0.2)

  45. 0.1M 0.15M 0.15M Initial -x M Change +x M +x M (0.1-x) M (0.15+x) M (0.15+x) M Equilibrium (0.15+x)2 [B][C] = Kc = (0.1-x) [A] Disturbing a System at Equilibrium: Calculations • Since Q<Kc the reaction will shift to the right to reestablish the equilibrium.(O<Kc, 反應向右以達成另一平衡) 2. Use algebra to represent the new concentrations A(g) B(g) + C(g) (a+b)2 =a2+2ab+b2 =0.45 0.045-0.45x=0.0225+0.30x+x2 x2+0.75x-0.0225=0

  46. Disturbing a System at Equilibrium: Calculations ax2+bx+c=0 x2+0.75x-0.0225=0 -b  b2-4ac x= 2a (0.075)2-4(1)(-0.0225) -0.75  x= 2x1 -0.75  0.081 x= -0.78 and 0.03M x= 2 Since 0<x<0.10  x=0.03M [A]=0.10-x=0.07 M [B]=[C]=0.15+x=0.18 M

  47. Instantaneous 0.40M 0.60M 0.60M [B][C] Q = [A] Disturbing a System at Equilibrium: Calculations Example 17-13: Refer to example 17-12. If the initial volume of the reaction vessel were halved, while the temperature remains constant, what will the new equilibrium concentrations be? Recall that the original concentrations were: [A]=0.20 M, [B]=0.30 M, and [C]=0.30 M. A(g) B(g) + C(g) 體積減半,濃度增倍 (0.6)(0.6) = =0.90 (0.40) Q>Kc thus the equilibrium shifts to the left or reactant side

  48. 0.40M 0.60M 0.60M Initial +x M Change -x M -x M (0.4+x) M (0.60-x) M (0.60-x) M Equilibrium (0.60-x)2 [B][C] = Kc = (0.4+x) [A] Disturbing a System at Equilibrium: Calculations • Set up the algebraic expressions to determine the equilibrium concentrations 反應向左 A(g) B(g) + C(g) 0.18+0.45x=0.36-1.2x+x2 x2-1.65x+0.18=0 =0.45 (-1.65)2-4(1)(0.18) 1.65  Since 0<x<0.60  x=0.12M [A]=0.40+x=0.52 M [B]=[C]=0.60-x=0.48 M x= 2x1 1.65  1.42 x= 2 x= 1.5 and 0.12M

  49. Example 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. Equilibrium Equilibrium 0.6/2 0.3M 0.1M 0.2/2 1.2/2 0.6M [COCl2] = [CO][Cl2] Disturbing a System at Equilibrium: Calculations CO(g)+ Cl2(g) COCl2(g) (0.6) = Kc =20 (0.30)(0.10)

  50. -x M Change -x M +x M Orig. Equil. 0.3M 0.1M 0.6M (Stress) Add +0.4M 0.30 M 0.50M 0.6M [COCl2] New Initial = [CO][Cl2] Equilibrium (0.3-x) M (0.50-x)M (0.60+x)M Disturbing a System at Equilibrium: Calculations An additional 0.80 mole of Cl2is added to the vessel at the same temperature. Calculate the molar concentrations of CO, Cl2, and COCl2 when the new equilibrium is established. 0.80 mole of Cl2 in 2-liter vessel  0.4M of Cl2 CO(g)+ Cl2(g) COCl2(g) Q<Kc 反應向右 (0.6+x) 20x2-17x+2.4=0 = Kc =20 (0.3-x)(0.5-x) Since 0<x<0.30  x=0.18M [CO]=0.30-x=0.12 M [Cl2]=0.5-x=0.32 [COCl]=0.6+x=0.78 (17)2-4(20)(2.3) 17  x= 2x20 (0.6) = =4 Qc X=0.67 and 0.18 (0.30)(0.50)

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