Chemical Equilibrium - Science14化學(xué)平衡的科學(xué)

1、16 Chemical Equilibrium116 Chemical EquilibriumReactions seldom go complete such that one of the reactants is exhausted as we have discussed in stoichiometry and limiting reagent. Instead, they are mostly reversible. For example,CO (g) + 3 H2 (g) CH4 (g) + H2O (g)The two-head arrow (or equal signs =

2、) indicates that the reaction goes both ways. Similarly,N2O4 (colorless gas) 2 NO2 (brown gas) (show a film)When the forward and reverse reaction rates are equal, the system is said to be at equilibrium.16 Chemical Equilibrium2 The NASA Computer program CEA (Chemical Equilibrium with Applications) c

3、alculates chemical equilibrium compositions and properties of complex mixtures. Applications include assigned thermodynamic states, theoretical rocket performance, Chapman-Jouguet detonations, and shock-tube parameters for incident and reflected shocks.16 Chemical Equilibrium31 mol CH42 mol S2T = 10

4、00 KP = 1.5 atmCountours show G/RTEquilibrium occur at minimum G/RT16 Chemical Equilibrium4A SystemScientific experiments usually investigate a system, which is isolated from its environment or surrounding.The system can be a nucleus, an atom, a molecule, a plant, an animal, an experimental setup, t

5、he Earth, the solar system, the galaxy, or the universe.Matter and energy can be transferred into or out of a system.The environment orThe surroundingThe open or closed system16 Chemical Equilibrium5Chemical Equilibrium a dynamic processChemical equilibrium is a state of a system when reaction rates

6、 in both directions are equal. Changes continue at the molecular level, but the macroscopic properties stay the same. Equilibrium is a dynamic rather than static process.Heat is energy flowing from a high temperature object to a low temperature object. Equilibrium is reached when the temperatures ar

7、e the same. Molecules still exchange kinetic energies during collision. Water flows from a high potential-energy place to a low potential-energy place. When potential energies are the same, water stops flowing. Water-molecule diffusion continues.16 Chemical Equilibrium6Some Typical EquilibriaMacrosc

8、opic properties such as vapor pressure, solubility, distribution of a substance in two immiscible liquids are typical equilibria.Vapor pressure of a liquid is the pressure of the vapor in equilibrium with a liquid. It is a function of temperature.Solubility, amount in solution in equilibrium with so

9、lid, depends on the substance and the solvent, often also on temperature.Distribution (or equilibrium) coefficients of a substance in immiscible liquids are useful guides for the separation of mixtures into components.16 Chemical Equilibrium7The reaction quotientsFor any chemical reaction in a syste

10、m, a A + b B c C + d Dwe can always write a quotient,Cc DdQ = (Q is reaction quotient)Aa Bbfor example, NO22Q = for N2O4 2 NO2 N2O4Know the definition of Q. Text: 16-5 defines Q & 16-1 defines K.16 Chemical Equilibrium8Expressions for reaction quotientsFor the ionization of an acid, H2SO4 = 2 H+

11、 + SO42-what is the reaction quotient expression? Hint: The reaction quotient is H+2 SO42- = Q H2SO4 If M is used for the concentrations, what are the units for Q? 16 Chemical Equilibrium9More expressions for reaction quotients2 H2 + O2 2 H2O H2O2Q = H22 O2 H22 O2 1 Q r = = H2O2 Q2 H2O 2 H2 + O24 H2

12、 + 2 O2 4 H2O H2O4 Q2 = = Q 2 H24 O22Q = Q 2 Note the relationshipsAll relations are due to definitions. See 16-316 Chemical Equilibrium10Equilibrium constant - mass action lawWhen a chemical reaction in a closed system is at equilibrium, the reaction quotient is defined as the equilibrium constant,

13、 K, which depends on T. For a chemical reaction in a closed system,a A + b B c C + d DthenCc DdQ = = K(at equilibrium)Aa BbUnits of K depends on units of s, M(a+b-c-d) if M is used.If Q K, the reaction goes backward | to decrease productSee 16-5 & 16-316 Chemical Equilibrium11Evaluating KThe equ

14、ilibrium mixture is found to contain 0.07, 0.11, 0.03 and 0.03 moles of CO, H2, CH4 and H2O respectively for the reaction:CO (g) + 3 H2 (g) CH4 (g) + H2O (g)What is the equilibrium constant, K?Solution CO (g) + 3 H2 (g) CH4 (g) + H2O (g)start0.07 0.110.030.03 (0.03) (0.03) K = = 9.7 (0.07) (0.11)3Wh

15、at governs the quantitative relationship among equilibrium mixtures?16 Chemical Equilibrium12Evaluate quantitiesWhen 0.10 moles CO and 0.20 moles of H2 are placed in a 1-L vessel at temperature T and allowed to come to equilibrium. The mixture is found to contain 0.03 moles H2O for the reaction:CO (

16、g) + 3 H2 (g) CH4 (g) + H2O (g)What is the equilibrium constant, K?Solution (note the strategy please)CO (g) + 3 H2 (g) CH4 (g) + H2O (g)start0.10 0.2000change - x - 3x+x+x (x=0.03)eqlbm 0.1-x0.2-3x0.030.03 0.070.110.030.03K = (0.03) (0.03) / (0.07) (0.11)3 = 9.716 Chemical Equilibrium13Evaluate qua

17、ntitiesWhen 0.08 moles CO, 0.14 moles of H2, and 0.08 moles of CH4 are placed in a 1-L vessel at temperature T and allowed to come to equilibrium. The mixture is found to contain 0.01 moles H2O for the reaction:CO (g) + 3 H2 (g) CH4 (g) + H2O (g)What is the equilibrium constant, K?Solution (note the

18、 strategy please)CO (g) + 3 H2 (g) CH4 (g) + H2O (g)start0.08 0.140.080change - x - 3x 0.08+x+x (x=0.01)eqlbm 0.08-0.01 0.14-0.03 0.08+0.01 0.01 0.07 0.110.090.01K = (0.09) (0.01) / (0.07) (0.11)3 = 9.716 Chemical Equilibrium14EntropyWe all know that heat flow from a high-T object into a low-T objec

19、t. This type of change can be described as due to increase of entropy.Entropy S is the amount of energy transferred, E, divided by T.ES = or T S = E TEntropy increases for spontaneous changes (reactions).Entropy is also associated with randomness of ordering.What causes change when no apparent energ

20、y is involved?16 Chemical Equilibrium15Entropy calculationA system consists 10.0 L of water at 300 K and 3.0 g of ice at 273 K. Heat of fusion of ice (molar mass 18) is 6 kJ/mol. What is the change in entropy when the system is at equilibrium.Solution:Since the amount of heat for melting the ice is

21、6 *(3/ 18) = 1.0 kJ too little to change the temperature of the entire system appreciably from 300 K. Thus the entropy change is 1/300 + 1/273 = 3.3e 4 kJ/K = 0.33 J/K (entropy unit)What is the molar entropy for ice melting?+6000 J mol1 / 273 K = 22 J mol 1 K 1 (entropy unit mol1)16 Chemical Equilib

22、rium16Josiah W. Gibbs(1839-1903) New Haven, Connecticut, USA First Doctorate (1863 Yale) in Engineering of U.S.A.Studied in Europe, influenced by Kirchhoff & Helmholtz.1871 Professor of mathematics and physics at YaleUnassuming in manner, genial and kindly in his intercourse with his fellow-men,

23、 never showing impatience or irritation, devoid of personal ambition of the baser sort or of the slightest desire to exalt himself, he went far toward realizing the ideal of the unselfish, Christian gentleman. In the minds of those who knew him, the greatness of his intellectual achievements will ne

24、ver overshadow the beauty and dignity of his life. - Bumsteadlunar crater Gibbs16 Chemical Equilibrium17Gibbs free energy, G or GFollowing the definition of entropy S, Gibbs defined a function together with enthalpy of change, H.G = H T S now called Gibbs free energy.orG = H T Sfor differences of th

25、ese quantitiesSince H is negative and S is positive for spontaneous change, Gibbs free energy is negative for all spontaneous changes, and it unified enthalpy and entropy for changes. Gibbs free energy is the maximum amount of available energy in any change.Similar to Ho, Go is the Gibbs free energy

26、 at the standard T and P.What drives physical and chemical changes?16 Chemical Equilibrium18Energy and equilibrium constant KA quantity called Gibbs free energy, G, is defined as the maximum amount of energy from a chemical reaction for doing work. G is negative for spontaneous reaction G is positiv

27、e for non-spontaneous reactionThe G is related to the reaction quotient Q, G = Go + R T ln QWhen a system is at equilibrium there is no available energy, G = Go + R T ln K = 0Thus,Go = R T ln K, ln K = Go / R TK = exp ( Go / R T)See 20-2 to 20-6 page 784-808 plse16 Chemical Equilibrium19More express

28、ions for equilibrium constant2 H2 + O2 2 H2O H2O2K = H22 O2 H22 O2 1 K r = = H2O 2 K2 H2O 2 H2 + O24 H2 + 2 O2 4 H2O H2O4 K2 = = K 2 H24 O22 K 2 = K Note these relationships in light of ln K = Go / R T 16 Chemical Equilibrium20An ExampleHydrogen iodide, HI, decomposes,2 HI H2 + I2The amount of I2 ca

29、n be determined from the intensity of the violet color of I2. When 4.00 mol of HI was placed in a 5.00-L vessel at 458oC, the equilibrium mixture was found to contain 0.442 mol I2. What is the value for K for the decomposition of HI at this temperature? See 16-4 and 16-7 for this and the following e

30、xamples.16 Chemical Equilibrium21Analysis of ExampleHydrogen iodide, HI, decomposes,2 HI H2 + I2The amount of I2 can be determined from the intensity of the violet color of I2. When 4.00 mol of HI was placed in a 5.00-L vessel at 458oC, the equilibrium mixture was found to contain 0.442 mol I2. What

31、 is the value for K for the decomposition of HI at this temperature? Analysis of the problem:HI0 = 4.0 mol / 5.00 L = 0.800 MI2equilibrium = 0.442 mol / 5.0 L = 0.0884 MKnow your strategy 16 Chemical Equilibrium22Calculate KcExample 14.3 The equilibrium concentration of various species are given bel

32、ow their symbols,2 HI (g) = H2 (g) + I2 (g)start0.800 00change = ? = ?= 0.0884 M from experiment equlbm0.800-2x xx value0.62320.08840.0884 M0.0884*0.0884 Kc = = 0.0201 (0.6232)2Skill: re-interpret question. Evaluate other quantities and K from known start conditions and experimental resultsEvaluate

33、Go16 Chemical Equilibrium23Evaluate KcCompare with slide 41At 300 K and 1 atm, an equilibrium mixture contains 70% NO2 by mass in a tube of NO2 and N2O4. Evaluate Kc for the reaction,N2O4 = 2 NO2The challenge (hint):assume 100 g total mass (70 g of NO2 and 30 g N2O4) find amount in mole for each (1.

34、522 mol NO2; 0.326 mol N2O4) find volume from pressure for all (45.49 L) find concentration from volume (0.0335 M NO2; 0.00717 M)evaluate Kc (equation dependent)16 Chemical Equilibrium24Application of KcFor N2O4 (g) = 2 NO2 (g) Kc = 4.20e-3 at 300 K,What are the equilibrium concentrations if 0.100 m

35、ol of N2O4 is placed in a 5.00-L flask?Strategy:write down the equationfigure out the initial condition assume the change to be xfigure out the equilibrium concentrationuse the concentration-Kc relationshipsolve for x and then check the results.N2O4 (g) = 2 NO2 (g)0.0200-x 2 x (2x)2 = 4e-3 (K) 0.020

36、0-x4x2 + 0.0042 x 0.0000926 = 0 -0.00463 + (4.63e-3)2 4*4*(-9.265e-5)x = = 4.27e-32*4N2O4 (g) = 2 NO2 (g)0.0157 8.54e-3 M16 Chemical Equilibrium25Evaluate equilibrium concentrations of all for the reaction condition as given in the start: (at T1)N2 (g) + 3 H2 (g) 2 NH3 (g)K = 1e-5start1.0 2.00change

37、 - x - 3x+2x (= ?) eqlbm 1.0-x2.0-3x 2x (2 x)2 = 1e-5(1-x)(2.0-3x)3 4x2 = 1e-5 (1.0-x)(2.0-3x)3x2 = 1e-5*1.0*2.03/4 = 2e-5x = 4.5e-3 (justified)ApproximationsResults:N2 = 0.995 MH2 = 1.99NH3 = 9e-3Checking(2*4.5e-3)2 = 1e-50.995*1.993Discuss16 Chemical Equilibrium26Trial and error or Newtons methodW

38、hen 0.10 moles CO and 0.20 moles of H2 are placed in a 1-L vessel at temperature T and allowed to come to equilibrium. CO (g) + 3 H2 (g) CH4 (g) + H2O (g) Kc = 9.7What are the concentrations of various species? Solution (note the strategy please)CO (g) + 3 H2 (g) CH4 (g) + H2O (g)start0.10 0.2000cha

39、nge - x - 3x+x+x eqlbm 0.1-x0.2-3x x x (= ?)Kc = (x)2 / (0.1-x) (0.2-3x)3 = 9.7Too difficult to solve.You need not worry.Trial-and-error (Newtons method) results shown x =Kc =0.05*4000.0615174 0.04 520.0311.30.0298.210.0299 9.50*x note = 7.09(0.01-x)2(0.01-x)16 Chemical Equilibrium28Successive appro

40、ximationFor 2 H2S (g) = 2 H2 (g) + S2 (g) Kc = 4.2e-6 at 831oCWhat is S2 when 0.0700 mol H2S is allowed to reach equilibrium in a 1.00-L vessel.Solution: 2 H2S (g) = 2 H2 (g) + S2 (g) 0.0700-2x2 x x (2x)2 x = 4.2e-6x = 3(4.2e-6*0.07002/4) = 1.73e-3(0.0700-2x)2 assume 0 (2x)2 x = 4.2e-6x = 3(4.2e-6*0

41、.06652/4) = 1.67e-3(0.0700-2*0.00173)2 successive inclusionCarry out the calculation plse4x3(0.07-2*0.00167) = 4.2e-6 x = 3(4.2e-6*0.06672/4) = 1.67e-316 Chemical Equilibrium29Equilibria involving solventsSince solvent is essentially constant, it does not appear in Kc expression. You just have to kn

42、ow and apply.CH3COOC2H5 (aq) + H2O (l) = CH3COOH (aq) + C2H5OH (aq) CH3COOH C2H5OH Kc = CH3COOC2H5 H2O CH3COOH C2H5OH Usually, Kc = Kc H2O = CH3COOC2H5 16 Chemical Equilibrium30Equilibria involving solvents - applicationFor the reaction,CH3COOC2H5 (aq) + H2O (l) = CH3COOH (aq) + C2H5OH (aq)Kc = 0.25

43、 at 25oC. Calculate C2H5OH at equilibrium starting with 1.00 M of ethyl acetate.Solution:CH3COOC2H5 (aq) + H2O (l) = CH3COOH (aq) + C2H5OH (aq)1.00 xxx= C2H5OH x2 + 0.25 x 0.25 = 0 0.25 + 0.252 4(-0.25) x = = 0.39 M 2 x2 = 0.25(1.00 x)16 Chemical Equilibrium31Equilibrium Calculations a summarySevera

44、l methods have been introduced in evaluating xSimple methodApproximationTrial and error (Newtons method)Recognize math relationship (taking square root on both sides)Successive approximation16 Chemical Equilibrium32Kp and Kc; Kp = Kc (R T) c+d-(a+b)The equilibrium constants depends on the units used

45、 for quantities. In general, the equilibrium constant is represented by KWhen concentrations are used, Kc is the notation;when partial pressures are used, Kp is usedSince PV = n R T,P = (n/V) R T = R T = (n / V) is the concentrationFor the reaction a A + b B c C + d DPCc PDd Cc D d (R T)c + d Kp = =

46、 = Kc (R T)c+d-(a+b)PAa PBb Aa B b (R T)a + bSee page 63516 Chemical Equilibrium33Kp more appropriate forH2O (l) = H2O (g) Kp = vapor pressure of waterH2O (s) = H2O (g) Kp = vapor pressure of iceAt the triple point, its obvious thatvapor pressure of ice = vapor pressure of waterif these are equal, t

47、he three phases are present. CaCO3 (s) = CaO + CO2 Kp = partial v. p. of CO216 Chemical Equilibrium34Kc and Kp conversionFor 2 H2S (g) = 2 H2 (g) + S2 (g) Kc = 4.2e-6 at 831oCWhat is Kp?Solution: Kp = Kc (R T)c+d (a+b)(recall or derive) = Kc (R T)(2+1 2) = 4.2e-6 *8.314 kJ mol 1 K 1 * (831+273) K =

48、3.9e-22 H2S (g) = 2 H2 (g) + S2 (g) Kp confirmation0.0667 3.34e-3 1.67e-3Kc = 4.2e-6 at 831oC612 kPa 30.7 15.3Kp = 3.9e-2 J mol 1 K 1 at 831oC6.04 atm 0.30 0.15 Kp = 3.7e-4 L atm mol 1 K 1 16 Chemical Equilibrium35Le Chateliers PrincipleThe system will restore its equilibrium after changing conditio

49、ns.An open bottle of water will dry eventually, because the partial pressure never reach equilibrium (removing vapor pressure).Increase pressure will increase the formation of NH3 inN2 (g) + 3 H2 (g) = 2 NH3 (g)Hof = 91.8 kJIncrease T will decrease the formation of NH3 in this exothermic reaction.Ho

50、w does CH4 change when pressure increase for an equilibrium of CO (g) + 3 H2 (g) CH4 (g) + H2O (g)?Increasing pressure tends to reduce the number of molecules.Study sections: 16.616 Chemical Equilibrium36Le Chateliers Principle cont.Possible changes:Temperature, T Increase T favors forward reaction

51、in an endothermic reaction but favors reverse reaction in an exothermic reactionPressure, P and volume V Pressure and volume are related. Decreases of volume increase the pressureAmounts of product and reactants Increase amounts of reactants, products or both increases P and V16 Chemical Equilibrium

52、37Henri Louis Le ChatelierScientist would be wise to refrain from using the mathematical equation unless he understands the theory that it represents, and can make a statement about the theory that does not consist just in reading the equation. L. Pauling(1850-1936)A mining engineer investigated str

53、ucture of metals and alloys, translated work of Gibbs into French stated Henricus Vant Hoffs (Nobel Prize 1901) theory inword andwrote on industrial efficiency and labor management relations16 Chemical Equilibrium38Effect of a catalystThere are two factors regarding the speed of reactions: kinetics

54、and thermodynamics. Kc and Kp are thermodynamic parameters, related to G o. Thus, the equilibrium constants are not affected by a catalyst.Catalyst only affect reaction constant k, which is related to the activation energy.EReactionthermodynamickineticSee page 643-64616 Chemical Equilibrium39Equilib

55、ria - SummaryApply stoichiometry to equilibrium mixture in a systemWrite reaction quotient and equilibrium constant expression. Rewrite Qs and Ks when the equations change.Predict direction of reaction by comparing Q and K.Evaluate Ks from some know conditions.Know how energies H and G are related t

56、o equilibrium constants.Use various methods to evaluate concentrations of an equilibrium system.Apply Le Chateliers principle to predict changes in P, , and T.Separate kinetic and thermodynamic factors of reactions.16 Chemical Equilibrium40Equilibrium review -1Recall thatG = Go + R T ln K = 0andGo = R T ln K, orln K = Go / R TSince Go refers to the standard condition (1 a