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Chapter 6

Chapter 6. Chemical Equilibrium. Chapter 6: Chemical Equilibrium. 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions Involving Pressures 6.4 The Concept of Activity 6.5 Heterogeneous Equilibria 6.6 Applications of the Equilibrium Constant

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Chapter 6

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  1. Chapter 6 Chemical Equilibrium

  2. Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions Involving Pressures 6.4 The Concept of Activity 6.5 Heterogeneous Equilibria 6.6 Applications of the Equilibrium Constant 6.7 Solving Equilibrium Problems 6.8 LeChatelier’s Principle 6.9 Equilibria Involving Real Gases

  3. Nitrogen dioxide shown immediately after expanding

  4. Figure 6.1: Reaction of 2NO2(g) and N2O4(g) over time in a closed vessel

  5. Reddish brown nitrogen dioxide, NO2 (g)

  6. Reaching Equilibrium on the Macroscopic and Molecular Level N2O4 (g) 2 NO2 (g) Colorless Brown

  7. The State of Equilibrium kfwd [NO2]2 krev [N2O4] = = Keq For the Nitrogen dioxide - dinitrogen tetroxide equilibrium: N2O4 (g, colorless) = 2 NO2 (g, brown) At equilibrium: ratefwd = raterev ratefwd = kfwd[N2O4] raterev = krev[NO2]2 kfwd[N2O4] = krev[NO2]2 1) Small k N2 (g) + O2 (g) 2 NO(g)K = 1 x 10 -30 2) Large k 2 CO(g) + O2 (g) 2 CO2 (g)K = 2.2 x 1022 3) Intermediate k 2 BrCl(g) Br2 (g) + Cl2 (g)K = 5

  8. Reaction Direction and the Relative Sizes of Q and K

  9. Initial and Equilibrium Concentrations for theN2O4-NO2 System at 100°C Initial Equilibrium Ratio [N2O4] [NO2] [N2O4] [NO2] [NO2]2 [N2O4] 0.1000 0.0000 0.0491 0.1018 0.211 0.0000 0.1000 0.0185 0.0627 0.212 0.0500 0.0500 0.0332 0.0837 0.211 0.0750 0.0250 0.0411 0.0930 0.210

  10. Figure 6.2: Changes in concentration with time for the reaction H2O(g) + CO(g) H2 (g) + CO2 (g)

  11. Molecular model: When equilibrium is reached, how many molecules of H2O, CO, H2, and CO2 are present?

  12. Figure 6.3: H2O and CO are mixed in equal numbers H2O(g) + CO(g) H2 (g) + CO2 (g)

  13. Figure 6.4: Changes with time in the rates of forward and reverse reactions H2O(g) + CO(g) H2 (g) + CO2 (g)

  14. Figure 6.5: Concentration profile for the reaction

  15. Like Example 6.1 (P 195) - I • The following equilibrium concentrations were observed for the • Reaction between CO and H2 to form CH4 and H2O at 927oC. • CO(g) + 3 H2 (g) = CH4 (g) + H2O(g) • [CO] = 0.613 mol/L [CH4] = 0.387 mol/L • [H2] = 1.839 mol/L [H2O] = 0.387 mol/L • Calculate the value of K at 927oC for this reaction. • Calculate the value of the equilibrium constant at 927oC for: • H2O(g) + CH4 (g) = CO(g) + 3 H2 (g) • Calculate the value of the equilibrium constant at 927oC for: • 1/3 CO(g) + H2 (g) = 1/3 CH4 (g) + 1/3 H2O(g) • Solution: • a) Given the equation above: [CH4] [H2O] (0.387 mol/L) (0.387 mol/L) K = = = ______L2/mol2 [CO] [H2]3 (0.613 mol/L) (1.839 mol/L)3

  16. Like Example 6.1 (P 195) - II b) Calculate the value of the equilibrium constant at 927oC for: H2O(g) + CH4 (g) = CO(g) + 3 H2 (g) [CO] [H2]3 (0.613 mol/L) (1.839 mol/L)3 K = = = 25.45 mol2/L2 [H2O] [CH4] (0.387 mol/L) (0.387 mol/L) This is the reciprocal of K: 1 1 K = = 25.45 mol2/L2 0.0393 L2/mol2 • Calculate the value of the equilibrium constant at 927oC for: • 1/3 CO(g) + H2 (g) = 1/3 CH4 (g) + 1/3 H2O(g) [H2O]1/3 [CH4]1/3 (0.387mol/L)1/3 (0.387 mol/L)1/3 K = = [CO]1/3 [H2] (0.613 mol/L)1/3 (1.839 mol/L) (0.729) (0.729) K = = 0.340 L2/3/mol2/3 = (0.0393L2/mol2)1/3 (0.850)(1.839)

  17. Summary: Some Characteristics of the Equilibrium Expression The equilibrium expression for a reaction written in reverse is the reciprocal of that for the original reaction. When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression raised to the nth power. Thus Knew = (Koriginal)n The apparent units for K are determined by the powers of the various concentration terms. The (apparent) units for K therefore depend on the reaction being considered. We will have more to say about the units for K in section 6.4.

  18. Expressing K with Pressure Units n V For gases, PV=nRT can be rearranged to give: P = RT nP VRT n V or: = Since = Molarity, and R is a constant if we keep the temperature constant then the molar concentration is directly proportional to the pressure. Therefore for an equilibrium between gaseous compounds we can express the reaction quotient in terms of partial pressures. For: 2 NO(g) + O2 (g) 2 NO2 (g) If there is no change in the number of moles of reactants and products then n = 0 then Kc = Kp , or if there is a change in the number of moles of reactants or products then: P 2NO2 Qp = P 2NO x PO2 Kp = Kc(RT)ngas

  19. Figure 6.6: Position of the equilibrium CaCO3 (s) CaO(s) + CO2 (g)

  20. Writing the Reaction Quotient or Mass-Action Expression a A + bB cC + dD [C]c [D]d Q = [A]a[B]b Q = mass-action expression or reaction quotient Product of the Product Concentrations Q = Product of the Reactant Concentrations For the general reaction: Example: The Haber process for ammonia production: N2 (g) + 3 H2 (g) 2 NH3 (g) [NH3]2 Q = [N2][H2]3

  21. Reaction Direction and the Relative Sizes of Q and K

  22. Writing the Reaction Quotient from the Balanced Equation Problem: Write the reaction quotient for each of the following reactions: (a) The thermal decomposition of potassium chlorate: KClO3 (s) = KCl(s) + O2 (g) (b) The combustion of butane in oxygen: C4H10 (g) + O2 (g) = CO2 (g) + H2O(g) Plan: We first balance the equations, then construct the reaction quotient as described by equation 17.4. Solution: (a) 2 KClO3 (s) 2 KCl(s) + 3 O2 (g)Qc = [KCl]2[O2]3 [KClO3]2 (b) 2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O(g) [CO2]8 [H2O]10 Qc = [C4H10]2 [O2]13

  23. Writing the Reaction Quotient for an Overall Reaction–I Problem: Oxygen gas combines with nitrogen gas in the internal combustion engine to produce nitric oxide, which when out in the atmosphere combines with additional oxygen to form nitrogen dioxide. (1) N2 (g) + O2 (g) 2 NO(g)Kc1 = 4.3 x 10-25 (2) 2 NO(g) + O2 (g) 2 NO2 (g)Kc2 = 6.4 x 109 (a) Show that the overall Qc for this reaction sequence is the same as the product of the Qc’s for the individual reactions. (b) Calculate Kc for the overall reaction. Plan: We first write the overall reaction by adding the two reactions together and write the Qc. We then multiply the individual Kc’s for the total K. (1) N2 (g) + O2 (g) 2 NO(g) (2) 2 NO(g) + O2 (g) 2 NO2 (g) overall: N2 (g) + 2 O2 (g) 2 NO2 (g)

  24. Writing the Reaction Quotient for an Overall Reaction–II (a) cont. [NO]2 Qc (overall) = [N2][O2]2 For the individual steps: [NO]2 (1) N2 (g) + O2 (g) 2 NO(g) Qc1 = [N2] [O2] [NO2]2 (2) 2 NO(g) + O2 (g) 2 NO2 (g)Qc2 = [NO]2 [O2] [NO]2 [NO2]2 [NO2]2 Qc1 x Qc2 = x = The same! [N2] [O2] [NO]2 [O2] [N2][O2]2 (b) K = Kc1 x Kc2 = (4.3 x 10-25)(6.4 x 109) = ______________

  25. [SO2]2[O2] 1 Qc(rev) = = [SO3]2 Qc(fwd) The Form of Q for a Forward and Reverse Reaction The production of sulfuric acid depends upon the conversion of sulfur dioxide to sulfuric trioxide before the sulfur trioxide is reacted with water to make the sulfuric acid. 2 SO2 (g) + O2 (g) 2 SO3 (g) [SO3]2 Qc(fwd) = [SO2]2[O2] For the reverse reaction: 2 SO3 (g) 2 SO2 (g) + O2 (g) at 1000K Kc(fwd) = 261 1 1 and: Kc(fwd) = = = _____________ Kc(rev) 261

  26. [C] [B] Q1 = ; Q2 = [A] [C] Ways of Expressing the Reaction Quotient, Q Form of Chemical Equation Form of Q Value of K [B]eq [B] Reference reaction: A B Q(ref) = K(ref) = Reverse reaction: B A Q = = K = Reaction as sum of two steps: [A] [A]eq 1 [A] 1 K(ref) Q(ref) [B] (1) A C Qoverall = Q1 x Q2 = Q(ref)Koverall = K1 x K2 = x = (2) C B [C] [B] [B] = K(ref) [A] [C] [A] Coefficients multiplied by nQ = Qn(ref)K = Kn(ref) Reaction with pure solid or Q’ = Q(ref)[A] = [B] K’ = K(ref)[A] = [B] liquid component, such as A(s)

  27. Example 6.2 (P 202) - I [NH3]02 [N2]0[H2]03 • For the synthesis of ammonia at 500oC, the equilibrium constant is • 6.0 x 10-2 L2/mol2. Predict the direction in which the system will • shift to reach equilibrium in each of the following cases. • [NH3]0 = 1.0 x 10-3 M; [N2]0= 1.0 x 10-5 M; [H2]0=2.0 x 10-3 M • [NH3]0 = 2.00 x 10-4 M; [N2]0= 1.50 x 10-5 M; [H2]0= 3.54 x 10-1 M • [NH3]0 = 1.0 x 10-4 M; [N2]0= 5.0 M; [H2]0= 1.0 x 10-2 M • Solution • a) First we calculate the Q: (1.0 x 10-3 mol/L)2 Q = = = ____________________ L2/mol2 (1.0 x 10-5 mol/L)(2.0 x 10-3 mol/L)3 Since K = 6.0 x 10-2 L2/mol2, Q is much greater than K. For the system to attain equilibrium, the concentrations of the products must be decreased and the concentrations of the reactants increased. The system will shift to the left:

  28. Example 6.2 (P 202) - II [NH3]02 [NH3]02 [N2]0[H2]03 [N2]0[H2]03 b) We calculate the value of Q: (2.00 x 10-4 mol/L)2 Q = = = 6.01 x 10-2 L2/mol2 (1.50 x 10-5 mol/L) (3.54 x 10-1 mol/L)3) In this case Q = K, so the system is at equilibrium. No shift will occur. c) The value of Q is: (1.0 x 10-4 mol/L)2 Q = = = _________________ L2/mol2 (5.0 mol/L) (1.0 x 10-2 mol/L)3 Here Q is less than K, so the system will shift to the right, attaining equilibrium by increasing the concentration of the product and decreasing the concentrations of the reactants. More Ammonia!

  29. Like Example 6.3 (P203-5) - I Look at the equilibrium example for the formation of Hydrogen Chloride gas from Hydrogen gas and Chlorine gas. Initially 4.000 mol of H2, and 4.000 mol of Cl2, are added to 2.000 mol of gaseous HCl in a 2.000 liter flask. H2 (g) + Cl2 (g) 2 HCl(g) K = 2.76 x 102 = [Cl2] = [H2] = 4.000mol/2.000L = 2.000M [HCl] = 2.000 mol/2.000L = 1.000M Initial Concentration Change Equilibrium Conc. (mol/L) (mol/L) (mol/L) [H2]o = 2.000M -x [H2] = 2.000-x [Cl2]o = 2.000M -x [Cl2] = 2.000-x [HCl]o = 1.000M +2x [HCl] = 1.000 + 2x [HCl]2 [H2] [Cl2]

  30. Like Example 6.3 (P203-5) - II [HCl]2 [H2] [Cl2] (1.000 + 2x)2 (2.000 –x)(2.000 – x) (1.000 + 2x)2 (2.000 – x)2 K = 2.76 x 102 = = = Take the square root of each side: 16.61 = (1.000 + 2x) (2.000 – x) 33.22 – 16.61x = 1.000 + 2x Therefore: [H2] = 0.269 M 32.22 = 18.61x [Cl2] = 0.269 M x = 1.731 [HCl] = 4.462 M Check: = = 276 OK! [HCl]2 [H2] [Cl2] (4.462)2 (0.269)(0.269)

  31. Summary: Solving Equilibrium Problems Write the balanced equation for the reaction. Write the equilibrium expression using the law of mass action. List the initial concentrations. Calculate Q and determine the direction of the shift to equilibrium. Determine the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown. Check your calculated equilibrium concentrations by making sure that they give the correct value of K.

  32. 0.250 mol [CS2] = 4.70 L Determining Equilibrium Concentrations from K–I Problem: One laboratory method of making methane is from carbon disulfide reacting with hydrogen gas, and Kthis reaction at 900°C is 27.8. CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2 S(g) At equilibrium the reaction mixture in a 4.70 L flask contains 0.250 mol CS2, 1.10 mol of H2, and 0.45 mol of H2S, how much methane was formed? Plan: Write the reaction quotient, and calculate the equilibrium concentrations from the moles given and the volume of the container. Use the reaction quotient and solve for the concentration of methane. Solution: CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2 S(g) [CH4] [H2S]2 K = = 27.8 [CS2] [H2]4 [CS2] = ____________ mol/L

  33. Determining Equilibrium Concentrations from K–II Solution cont. 1.10 mol 0.450 mol [H2] = = 0.23404 mol/L [H2S] = = 0.095745 mol/L 4.70 L 4.70 L Kc [CS2] [H2]4 (27.8)(0.05319)(0.23404)4 [CH4] = = [H2S]2 (0.095745)2 0.004436 [CH4] = = 0.485547 mol/L = 0.486 M 0.009167 Check: Substitute the concentrations back into the equation for K and make sure that you get the correct value of K (0.485547 M)(0.095745 M)2 [CH4] [H2S]2 K = = = 27.81875 [CS2] [H2]4 (0.05319 M)(0.23404 M)4 OK!

  34. Determining Equilibrium Concentrations from Initial Concentrations and K –I [HF]2 K = = 115 [H2] [F2] Problem: Given the that the reaction to form HF from molecular hydrogenand fluorine has a reaction quotient of 115 at a certain temperature. If 3.000 mol of each component is added to a 1.500 L flask, calculate the equilibrium concentrations of each species. H2 (g) + F2 (g) 2 HF(g) Plan: Calculate the concentrations of each component, and then figure the changes, and solve the equilibrium equation to find the resultant concentrations. Solution: 3.000 mol [H2] = = 2.000 M 1.500 L 3.000 mol [F2] = = 2.000 M 1.500 L 3.000 mol [HF] = = 2.000 M 1.500 L

  35. [HF]2 K = = 115 = = [H2][F2] [HF]2 K = = [H2][F2] Determining Equilibrium Concentrations from Initial Concentrations and K–II Concentration (M) H2 F2 HF Initial 2.000 2.000 2.000 Change -x -x +2x Final 2.000-x 2.000-x 2.000+2x (2.000 + 2x)2 (2.000 + 2x)2 (2.000 - x) (2.000 - x) (2.000 - x)2 Taking the square root of each side we get: (2.000 + 2x) (115)1/2 = =10.7238 x = 1.528 (2.000 - x) (5.056 M)2 [H2] = 2.000 - 1.528 = 0.472 M (0.472 M)(0.472 M) [F2] = 2.000 - 1.528 = 0.472 M check: K = 115 [HF] = 2.000 + 2(1.528) = 5.056 M

  36. Calculating K from Concentration Data–I Problem: Hydrogen iodide decomposes at moderate temperatures by the reaction below: When 4.00 mol HI was placed in a 5.00 L vessel at 458°C, the equilibrium mixture was found to contain 0.442 mol I2. What is the value of Kc ? Plan: First we calculate the molar concentrations, and then put them into the equilibrium expression to find it’s value. Solution: To calculate the concentrations of HI and I2, we divide the amounts of these compounds by the volume of the vessel. 2 HI(g) H2 (g) + I2 (g) 4.00 mol 5.00 L Starting conc. of HI = = 0.800 M 0.442 mol 5.00 L Equilibrium conc. of I2 = = 0.0884 M Conc. (M) 2HI(g) H2 (g) I2 (g) Starting 0.800 0 0 Change - 2x x x Equilibrium 0.800 - 2x x x = 0.0884

  37. Calculating K from Concentration Data–II [HI] = M = (0.800 - 2 x 0.0884) M = 0.623 M [H2] = x = 0.0884 M = [I2] [H2] [I2] ( 0.0884)(0.0884) Kc = = = ____________ (0.623)2 [HI]2 Therefore the equilibrium constant for the decomposition of Hydrogen Iodide at 458°C is only 0.0201 meaning that the decomposition does not proceed very far under these temperature conditions. We were given the initial concentrations, and that of one at equilibrium, and found the others that were needed to calculate the equilibrium constant.

  38. - b + b2 - 4ac 2a x = Using the Quadratic Formula to Solve for the Unknown Given the Reaction between CO and H2O: Concentration (M) CO(g) + H2O(g) CO2(g) + H2(g) Initial 2.00 1.00 0 0 Change -x -x +x +x Equilibrium 2.00-x 1.00-x x x [CO2][H2] (x) (x) x2 Qc = = = = 1.56 [CO][H2O] (2.00-x)(1.00-x) x2 - 3.00x + 2.00 We rearrange the equation: 0.56x2 - 4.68x + 3.12 = 0 ax2 + bx + c = 0 quadratic equation: [CO] = 1.27 M [H2O] = 0.27 M [CO2] = 0.73 M [H2] = 0.73 M 4.68 + (-4.68)2 - 4(0.56)(3.12) x = = 7.6 M and 0.73 M 2(0.56)

  39. Predicting Reaction Direction and Calculating Equilibrium Concentrations –I CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g) Problem: Two components of natural gas can react according to the following chemical equation: In an experiment, 1.00 mol CH4, 1.00 mol CS2, 2.00 mol H2S, and 2.00 mol H2 are mixed in a 250 mL vessel at 960°C. At this temperature, K = 0.036. (a) In which direction will the reaction go? (b) If [CH4] = 5.56 M at equilibrium, what are the concentrations of the other substances? Plan: The find the direction, we calculate Qc using the calculated concentrations from the data given, and compare it with Kc. (b) Based upon (a), we determine the sign of each component for the reaction table and then use the given [CH4] at equilibrium to determine the others. Solution: [H2S] = 8.00 M, [CS2] = 4.00 M and [H2 ] = 8.00 M 1.00 mol 0.250 L [CH4] = = 4.00 M

  40. Predicting Reaction Direction and Calculating Equilibrium Concentrations –II [CS2] [H2]4 4.00 x (8.00)4 Q = = = 64.0 [CH4] [H2S]2 4.00 x (8.00)2 Comparing Q and K: Q > K (64.0 > 0.036, so the reaction goes to the left. Therefore, reactants increase and products decrease their concentrations. (b) Setting up the reaction table, with x = [CS2] that reacts, which equals the [CH4] that forms. Concentration (M) CH4 (g) + 2 H2S(g) CS2(g) + 4 H2(g) Initial 4.00 8.00 4.00 8.00 Change +x +2x -x - 4x Equilibrium 4.00 + x 8.00 + 2x 4.00 - x 8.00 -4x Solving for x at equilibrium: [CH4] = 5.56 M = 4.00 M + x x = ____________ M

  41. Predicting Reaction Direction and Calculating Equilibrium Concentrations –III x = 1.56 M = [CH4] Therefore: [H2S] = 8.00 M + 2x = 8.00 M + 2(1.56 M) = _________ M [CS2] = 4.00 M - x = 4.00 M - 1.56 M = __________ M [H2] = 8.00 M - 4x = 8.00 M - 4(1.56 M) = __________ M [CH4] = __________ M

  42. Le Chatelier’s Principle “If a change in conditions (a “stress”) is imposed on a system at equilibrium, the equilibrium position will shift in a direction that tends to reduce that change in conditions.” A + B C + D + Energy For example: In the reaction above, if more A or B is added you will force the reaction to produce more product, if they are removed, it will force the equilibrium to form more reactants. If C or D is added you will force the reaction to form more reactants, if they are Removed from the reaction mixture, it will force the equilibrium to Form more products. If it is heated, you will get more reactants, and if cooled, more products.

  43. Henri Louis Le Chatelier Source: Photo Researchers

  44. Blue Anhydrous cobalt(II) chloride CoCl2 (s) + 6 H2O(g) CoCl2 6 H2O(s)

  45. Figure 6.7: Equilibrium mixture

  46. The Effect of a Change in Concentration–I CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) Add NH3 CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) Given an equilibrium equation such as : If one adds ammonia to the reaction mixture at equilibrium, it will force the reaction to go to the right producing more product. Likewise, if one takes ammonia from the equilibrium mixture, it will force the reaction back to produce more reactants by recombining H2 and HCN to give more of the initial reactants, CH4 and NH3. Forces equilibrium to produce more product. Remove NH3 Forces the reaction equilibrium to go back to the left and produce more of the reactants.

  47. The Effect of a Change in Concentration–II CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) If to this same equilibrium mixture one decides to add one of the products to the equilibrium mixture, it will force the equilibrium back toward the reactant side and increase the concentrations of reactants. Likewise, if one takes away some of the hydrogen or hydrogen cyanide from the product side, it will force the equilibrium to replace it. Forces equilibrium to go toward the reactant direction. Add H2 Remove HCN Forces equilibrium to make more produce and replace the lost HCN.

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