无机化学课件：12 Ch15 Ch16 Ch17 Equilibrium

1、Chemistry: Atoms FirstJulia Burdge & Jason OverbyChapter 15Chemical EquilibriumChemical Equilibrium1515.1 The Concept of Equilibrium15.2 The Equilibrium ConstantCalculating Equilibrium ConstantsMagnitude of the Equilibrium Constant15.3 Equilibrium ExpressionsHeterogeneous EquilibriaManipulating Equi

2、librium ExpressionsGaseous Equilibria15.4 Using Equilibrium Expressions to Solve ProblemsPredicting the Direction of a ReactionCalculating Equilibrium Concentrations15.5 Factors that Affect Chemical EquilibriumAddition or Removal of a SubstanceChanges in Volume and PressureChanges in TemperatureCata

3、lysisThe Concept of EquilibriumThe decomposition of N2O4 is a reversible process, meaning the products of the reaction can react to form reactants.The system is in equilibrium when the rates of the forward reaction and the reverse reaction are the same.rate forward = kfN2O4 and rate reverse = krNO22

4、15.1N2O4(g) 2NO2(g)The Concept of EquilibriumStarting with N2O4The Concept of EquilibriumStarting with NO2The Concept of EquilibriumSome important things to remember about equilibrium are:Equilibrium is a dynamic stateboth the forward and reverse reactions continue to occur, although there is no net

5、 change in reactant and product concentration over time.At equilibrium, the rates of the forward and reverse reactions are equal.Equilibrium can be established starting with only reactants, with only products, or with any mixture of reactants and products.The Equilibrium Constantrate forward = rate

6、reversekfN2O4eq = krNO22eqThe subscript “eq” denotes a concentration at equilibrium.RearrangingThe ratio of two constants (kf/kr) is also a constant:15.2equilibrium expressionN2O4(g) 2NO2(g)equilibrium constantThe Equilibrium ConstantNote the relationship between the equilibrium constant and the bal

7、anced chemical equation:N2O4(g) 2NO2(g)The Equilibrium ConstantThe reaction quotient (Qc ) is a fraction with product concentrations in the numerator and reactant concentrations in the denominator.Each concentration is raised to a power equal to the corresponding stoichiometric coefficient in the ba

8、lanced chemical equation.aA + bB cC + dDThis law of mass action applies to not only elementary reactions, but also to more complex reactions.(at equilibrium) Worked Example 15.1Strategy Use the law of mass action to write the equilibrium expression and plug in the equilibrium concentrations of all t

9、hree species to evaluate Kc.Carbonyl chloride (COCl2), also called phosgene, is a highly poisonous gas that was used on the battlefield of World War I. It is produced by the reaction of carbon monoxide with chlorine gas:CO(g) + Cl2(g) COCl2(g)In an experiment conducted at 74C, the equilibrium concen

10、trations of the species involved in the reaction were as follows: CO = 1.210-2 M, Cl2 = 0.054 M, and COCl2 = 0.14 M. (a) Write the equilibrium expression, and (b) determine the value of the equilibrium constant for this reaction at 74C.Solution (a) Kc = (b) Kc = = 216 or 2.2102COCl2COCl2(0.14)(1.210

11、-2)(0.054)Think About It When putting the equilibrium concentrations into the equilibrium expression, we leave out the units. It is common practice to express equilibrium constants without units. Worked Example 15.2Strategy Use the law of mass action to write reaction quotients.Write reaction quotie

12、nts for the following reactions:(a) N2(g) + 3H2(g) 2NH3(g)(b) H2(g) + I2(g) 2HI(g)(c) Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)(d) 2O3(g) 3O2(g)(e) Cd2+(aq) + 4Br-(aq) CdBr42-(aq)(f) 2NO(g) + O2(g) 2NO2(g)Solution (a) Qc = (d) Qc =NH32N2H23(b) Qc = HI2H2I2(c) Qc = Ag(NH3)2+Ag+NH32O23O32(e) Qc = CdBr42-Cd2+Br

13、-4(f) Qc = NO22NO2O2Think About It With practice, writing reaction quotients becomes second nature. Without sufficient practice, it will seem inordinately difficult. It is important that you become proficient at this. It is very often the first step in solving equilibrium problems.The Equilibrium Co

14、nstantAt any point during the progress of a reaction:aA + bB cC + dDThe Equilibrium ConstantThe value of the reaction quotient, Q, changes as the reaction progressesN2O4(g) 2NO2(g)The Equilibrium ConstantThe equilibrium constant gives the extent a reaction will proceed at a particular temperature.Th

15、ree outcomes are possible:1)The reaction will go essentially to completion and the equilibrium mixture will consist predominately of products.Ag+(aq) + 2NH3(aq) Ag(NH3)2+(g)Kc = 1.5 x 107 (at 25C)Large Kc, product favoredThe Equilibrium ConstantThe equilibrium constant gives the extent a reaction wi

16、ll proceed at a particular temperature.Three outcomes are possible:2)The reaction will not occur to any significant degree, and the equilibrium mixture will consist predominantly of reactant.N2(g) + O2(g) 2NO(g)Kc = 4.3 x 1025 (at 25C)3)The reaction will proceed a significant degree but will not go

17、to completion, and the equilibrium mixture will contain comparable amounts of both reactants and products.Small Kc, reactant favoredEquilibrium ExpressionsWhen the species in a reversible chemical reaction are not all in the same phase, the equilibrium is heterogeneous.Only gaseous species and aqueo

18、us species appear in equilibrium expressions, pure solids and pure liquids do not.15.3CO2(g) + C(s) 2CO(g)2Fe(s) + 3H2O(l) Fe2O3(s) + 2H2(g) Worked Example 15.3Strategy Use the law of mass action to write the equilibrium expressions for each reaction. Only gases and aqueous species appear in the exp

19、ression.Write equilibrium expressions for each of the following reactions:(a) CaCO3(s) CaO(s) + CO2(g)(b) Hg(l) + Hg2+(aq) Hg22+(aq)(c) 2Fe(s) + 3H2O(l) Fe2O3(s) + 2H2(g)(d) O2(g) + 2H2(g) 2H2O(l)Solution (a) Kc = CO2 (b) Kc = Hg22+Hg2+(d) Kc = 1O2H22(c) Kc = H22Think About It Like writing equilibri

20、um expressions for homogeneous equilibria, writing equilibrium expressions for heterogeneous equilibria becomes second nature if you practice. The importance of developing this skill now cannot be overstated. Your ability to understand the principles and to solve many of the problems in and Chapters

21、 16 to 19 depends on your ability to write equilibrium expressions correctly and easily.Equilibrium ExpressionsWhen a reversible chemical equation is manipulated, it is also necessary to make appropriate changes in the equilibrium expression and the equilibrium constant. Worked Example 15.4Strategy

22、Begin by writing the equilibrium expressions for the reactions that are given. Then, determine the relationship of each equations equilibrium expression to the equilibrium expression of the original equations, and make the corresponding change to the equilibrium constant for each.Kc =and Kc =The fol

23、lowing reactions have the indicated equilibrium constants at 100C:(1) 2NOBr(g) 2NO(g) + Br2(g)Kc = 0.014(2) Br2(g) + Cl2(g) 2BrCl(g)Kc = 7.2Determine the value of Kc for the following reactions at 100C:(a) 2NO(g) + Br2(g) 2NOBr(g) (d) 2NOBr(g) + Cl2(g) 2NO(g) + 2BrCl(g)(b) 4NOBr(g) 4NO(g) + 2Br2(g)

24、(e) NO(g) + BrCl(g) NOBr(g) + Cl2(g)(c) NOBr(g) NO(g) + Br2(g)1212NO2Br2NOBr2BrCl2Br2Cl2 Worked Example 15.4 (cont.)Solution (a) This equation is the reverse of original equation 1. Its equilibrium expression is the reciprocal of that for the original equation:Kc = = 1/0.014 = 71(b) This is original

25、 equation 1 multiplied by a factor of 2. Its equilibrium expression is the original expression squared:Kc = = (0.014)2 = 2.010-4(c) This is the original equation 1 multiplied by . Its equilibrium expression is the square root of the originalKc = or Kc = = (0.014)1/2 = 0.12NOBr2NO2Br2NO2Br2NOBr2212NO

26、2Br2NOBr21/2 Worked Example 15.4 (cont.)Solution (d) This is the sum of the original equations 1 and 2. Its equilibrium expression is the product of the two individual expressions:Kc = = (0.014)(7.2) = 0.10(e) Probably the simplest way to analyze this reaction is to recognize that it is the reverse

27、of the reaction in part (d), multiplied by . Its equilibrium expression is the square root of the reciprocal of the expression in part (d):Kc = = (1/0.10)1/2 = 3.2NO2BrCl2NOBr2Cl2NO2BrCl2NOBr2Cl21/212 Worked Example 15.4 (cont.)Think About It The magnitude of an equilibrium constant reveals whether

28、products or reactants are favored, so the reciprocal relationship between Kc values of forward and reverse reactions should make sense. A very large Kc value means that products are favored. In the reaction of hydrogen ion and hydroxide ion to form water, the value of Kc is very large, indicating th

29、at the product, water, is favored.H+(aq) + OH-(aq) H2O(l)Kc = 1.01014 (at 25C)Simply writing the equation backward doesnt change the fact that water is the predominate species. In the reverse reaction, therefore, the favored species is on the reactant side:H2O(l) H+(aq) + OH-(aq) Kc = 1.010-14 (at 2

30、5C)As a result, the magnitude of Kc should correspond to reactants being favored; that is, it should be very small.Gaseous EquilibriaWhen an equilibrium expression contains only gases, we can write an alternate form of the expression in which the concentrations of gases are expressed as partial pres

31、sures (atm). Thus, for the equilibriumwe can write asKc = orKP =The relationship between Kc and KP can be expressed aswhere n = moles of gaseous products moles of gaseous reactants.N2O4(g) 2NO2(g)NO22N2O4(PNO2)2PN2O4KP = Kc(0.08206 Latm/Kmol)Tn Worked Example 15.5Strategy Write equilibrium expressio

32、ns for each equation, expressing the concentrations of the gases in partial pressures.Write KP expressions for (a) PCl3(g) + Cl2(g) PCl5(g), (b) O2(g) + 2H2(g) 2H2O(l), and (c) F2(g) + H2(g) 2HF(g).Solution (a) All the species in this equation are gases, so they will all appear in the KP expression.

33、KP =(b) Only the reactants are gases. KP =(c) All species are gases. KP = (PPCl5)(PPCl3)(PCl2)1(PO2)(PH2)2(PHF)2(PF2)(PH2)Think About It It isnt necessary for every species in the reaction to be a gasonly those species that appear in the equilibrium expression. Worked Example 15.6Strategy Use KP = K

34、c(0.08206 Latm/Kmol)Tn. Be sure to convert temperature in degrees Celsius to kelvins. Using n = moles of gaseous products moles of gaseous reactants, n = 2(NO2) 1(N2O4) = 1. T = 298K.The equilibrium constant, Kc, for the reactionN2O4(g) 2NO2(g)is 4.6310-3 at 25C. What is the value of KP at this temp

35、erature.Solution KP = Kc T = (4.6310-3)(0.08206 298) = 0.1130.08206 LatmKmolThink About It Note that we have essentially disregarded the units of R and T so that the resulting equilibrium constant, KP, is unitless. Equilibrium constants commonly are treated as unitless quantities.Using Equilibrium E

36、xpressions to Solve ProblemsThe equilibrium expression may be used to predict the direction of a reaction and to calculate equilibrium concentrations.Predictions are made based on comparisons between Qc and Kc.There are three possibilities: Q KThe ratio of initial concentrations of products to react

37、ants is too large. To reach equilibrium products must be converted to reactants. The system proceeds in the reverse direction.15.4 Worked Example 15.7Strategy Use the initial concentrations to calculate Qc, and then compare Qc with Kc.Qc = = = 0.61At 375C, the equilibrium constant for the reactionN2

38、(g) + 3H2(g) 2NH3(g)is 1.2. At the state of a reaction, the concentrations of N2, H2, and NH3 are 0.071 M, 9.210-3 M, and 1.8310-4 M, respectively. Determine whether the system is at equilibrium, and if not, determine in which direction it must proceed to establish equilibrium.NH3i2N2iH2i3(1.8310-4)

39、2(0.071)(9.210-3)3Strategy The calculated value of Qc is less than Kc. Therefore, the reaction is not at equilibrium and must proceed to the right to establish equilibrium.Think About It In proceeding to the right, a reaction consumes reactants and produces more products. This increases the numerato

40、r in the reaction quotient and decreases the denominator. The result is an increase in Qc until it is equal to Kc, at which point equilibrium will be established.Calculating Equilibrium ConcentrationsEquilibrium concentrations can be calculated from initial concentrations if the equilibrium constant

41、 is known.Kc = 24.0 (200C)Initial concentration (M)0.8500Change in concentration (M)Equilibrium concentration (M)x0.850 x+xxcis-Stilbenetrans-Stilbenecis-Stilbenetrans-Stilbene Using Equilibrium Expressions to Solve ProblemsInitial concentration (M)0.8500Change in concentration (M)Equilibrium concen

42、tration (M)x0.850 x+xxcis-Stilbenetrans-Stilbene Use the equilibrium concentrations, defined in terms of x, in the equilibrium expression:x = 0.816 M (Represents the change in initial concentrations)Using Equilibrium Expressions to Solve ProblemsInitial concentration (M)0.8500Change in concentration

43、 (M)Equilibrium concentration (M)0.8160.850 0.816+0.8160.816cis-Stilbenetrans-Stilbene Calculate the equilibrium concentrations of cis- and trans-stilbene:cis-stilbene = (0.850 x) M = (0.850 0.816) M = 0.034 Mtrans-stilbene = x M = 0.816 M Worked Example 15.8Strategy Insert the starting concentratio

44、ns that we know into the equilibrium table:Kc for the reaction of hydrogen and iodine to produce hydrogen iodide,H2(g) + I2(g) 2HI(g)is 54.3 at 430C. What will the concentrations be at equilibrium if we start with 0.240 M concentrations of both H2 and I2?Initial concentration (M)0.2400.2400Change in

45、 concentration (M)Equilibrium concentration (M)H2I22HI+ Worked Example 15.8 (cont.)Solution We define the change in concentration of one of the reactants as x. Because there is no product at the start of the reaction, the reactant concentration must decrease; that is, this reaction must proceed in t

46、he forward direction to reach equilibrium. According to the stoichiometry of the chemical reaction, the reactant concentrations will both decrease by the same amount (x), and the product concentration will increase by twice that amount (2x). Combining the initial concentration and the change in conc

47、entration for each species, we get expressions (in terms of x) for the equilibrium concentrations.Initial concentration (M)0.2400.2400Change in concentration (M)xx+2xEquilibrium concentration (M)0.240 x0.240 x2xH2I22HI+ Worked Example 15.8 (cont.)Solution Next, we insert these expressions for the eq

48、uilibrium concentrations into the equilibrium expression and solve for x.Kc =54.3 = = = x = 0.189Use the calculated value of x, we can determine the equilibrium concentration of each species as follows:H2 = (0.240 x) M = 0.051 MI2 = (0.240 x) M = 0.051 MHI = 2x = 0.378 MHI2H2I2(2x)2(0.240 x)(0.240 x

49、)(2x)2(0.240 x)22x0.240 xThink About It Always check your answer by inserting the calculated concentrations into the equilibrium expression:The small difference between the calculated Kc and the one given in the problem statement is due to rounding.= 54.9 KcHI2H2I2(0.378)2(0.051)2= Worked Example 15

50、.9Strategy Using the initial concentrations, calculate the reaction quotient, Qc, and compare it to the value of Kc (given in the problem statement of Worked Example 15.8) to determine which direction the reaction will proceed to establish equilibrium. Then, construct an equilibrium table to determi

51、ne the equilibrium concentrations.For the same reaction and temperature as in Worked Example 15.8, calculate the equilibrium concentrations of all three species if the starting concentrations are as follows: H2 = 0.00623 M, I2 = 0.00414 M, HI = 0.0424 M.HI2H2I2(0.0424)2(0.00623)(0.00414)= 69.7 Worke

52、d Example 15.9 (cont.)Strategy Therefore, Qc Kc, so the system will proceed to the left (reverse) to reach equilibrium. The equilibrium table isInitial concentration (M)0.006230.004140.0424Change in concentration (M)Equilibrium concentration (M)H2I22HI+ Worked Example 15.9 (cont.)Solution Because we

53、 know the reaction must proceed from right to left, we know that the concentration of HI will decrease and the concentrations of H2 and I2 will increase. Therefore, the table should be filled in as follows:Next, we insert these expressions for the equilibrium concentration into the equilibrium expre

54、ssion and solve for x.Initial concentration (M)0.006230.004140.0424Change in concentration (M)+x+x2xEquilibrium concentration (M)0.00623 + x0.00414 + x0.0424 2xH2I22HI+HI2H2I2Kc =(0.0424 2x)2(0.00623 + x)(0.00414 + x)54.3 = Worked Example 15.9 (cont.)Solution It isnt possible to solve this equation

55、the way we did in Worked Example 15.8 (by taking the square root of both sides) because the concentrations of H2 and I2 are unequal. Instead, we have to carry out the multiplications.54.3(2.5810-5 + 1.0410-2x + x2) = 1.8010-3 1.7010-1x + 4x2Collecting terms we get50.3x2 + 0.735x 4.0010-4 = 0This is

56、a quadratic of the form ax2 + bx + c = 0. The solution for the quadratic equation Appendix 1 isx = Here we have a = 50.3, b = 0.735, and c = -4.0010-4, sox = Worked Example 15.9 (cont.)Solution x = 5.2510-4orx = 0.0151Only the first of these values, 5.2510-4, makes sense because concentration cannot

57、 be a negative number. Using the calculated value of x, we can determine the equilibrium concentration of each species as follows:H2 = (0.00623 + x) M = 0.00676 MI2 = (0.00414 + x) M = 0.00467 MHI = (0.0424 2x) M = 0.0414 MThink About It Checking this result givesHI2H2I2(0.0414)2(0.00676)(0.00467)=

58、54.3Kc = Worked Example 15.10Strategy Construct an equilibrium table to determine the equilibrium partial pressures.A mixture of 5.75 atm of H2 and 5.75 atm of I2 is contained in a 1.0-L vessel at 430C. The equilibrium constant (KP) for the reactionH2(g) + I2(g) 2HI(g)at this temperature is 54.3. De

59、termine the equilibrium partial pressures of H2, I2, and HI.Initial partial pressure (atm)5.755.750Change in partial pressure (atm)x x+2xEquilibrium partial pressure (atm)5.75 x5.75 x2xH2I22HI+ Worked Example 15.10 (cont.)Solution Setting the equilibrium expression equal to KP,54.3 = Taking the squa

60、re root of both sides of the equation gives =The equilibrium partial pressures are PH2 = PI2 = 5.75 4.52 = 1.23 atm, and PHI = 9.04 atm.(2x)2(5.75 x)22x5.75 x2x5.75 x7.369(5.75 x) = 2x42.37 7.369x = 2x42.37 = 9.369xx = 4.527.369 =Think About It Plugging the calculated partial pressures into the equi