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Chapter 14 Chemical Equilibrium

Chapter 14 Chemical Equilibrium. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. Arrow Conventions. Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions

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Chapter 14 Chemical Equilibrium

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  1. Chapter 14ChemicalEquilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

  2. Arrow Conventions • Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions • A single arrow indicates all the reactant molecules are converted to product molecules at the end • A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products •  in these notes Tro: Chemistry: A Molecular Approach, 2/e

  3. Reaction Dynamics • When a reaction starts, the reactants are consumed and products are made • forward reaction = reactants  products • therefore the reactant concentrations decrease and the product concentrations increase • as reactant concentration decreases, the forward reaction rate decreases • Eventually, the products can react to re-form some of the reactants • reverse reaction = products  reactants • assuming the products are not allowed to escape • as product concentration increases, the reverse reaction rate increases • Processes that proceed in both the forward and reverse direction are said to be reversible • reactants Û products Tro: Chemistry: A Molecular Approach, 2/e

  4. Hypothetical Reaction2 RedÛBlue The reaction slows over time, but the Red molecules never run out! At some time between 100 and 110 sec, the concentrations of both the Red and the Blue molecules no longer change – equilibrium has been established. Notice that equilibrium does not mean that the concentrations are equal! Once equilibrium is established, the rate of Red molecules turning into Blue is the same as the rate of Blue molecules turning into Red Tro: Chemistry: A Molecular Approach, 2/e

  5. Hypothetical Reaction2 RedÛBlue Tro: Chemistry: A Molecular Approach, 2/e

  6. Rate Forward Rate Reverse Reaction Dynamics Once equilibrium is established, the forward and reverse reactions proceed at the same rate, so the concentrations of all materials stay constant Eventually, the reaction proceeds in the reverse direction as fast as it proceeds in the forward direction. At this time equilibrium is established. Because the reactant concentration decreases, the forward reaction slows. As the products accumulate, the reverse reaction speeds up. As the forward reaction proceeds it makes products and uses reactants Initially, only the forward reaction takes place Rate Time Tro: Chemistry: A Molecular Approach, 2/e

  7. Dynamic Equilibrium • As the forward reaction slows and the reverse reaction accelerates, eventually they reach the same rate • Dynamic equilibriumis the condition wherein the rates of the forward and reverse reactions are equal • Once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant • because the chemicals are being consumed and made at the same rate Tro: Chemistry: A Molecular Approach, 2/e

  8. Equilibrium  Equal • The rates of the forward and reverse reactions are equal at equilibrium • But that does not mean the concentrations of reactants and products are equal • Some reactions reach equilibrium only after almost all the reactant molecules are consumed – we say the position of equilibrium favors the products • Other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed – we say the position of equilibrium favors the reactants Tro: Chemistry: A Molecular Approach, 2/e

  9. However, after a time, emigration will occur in both directions at the same rate, leading to populations in Country A and Country B that are constant, though not necessarily equal When Country A citizens feel overcrowded, some will emigrate to Country B An Analogy: Population Changes Tro: Chemistry: A Molecular Approach, 2/e

  10. Equilibrium Constant • Even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them • The relationship between the chemical equation and the concentrations of reactants and products is called the Law of Mass Action • For the general equation aA + bB ÛcC + dD, the Law of Mass Action gives the relationship below • the lowercase letters represent the coefficients of the balanced chemical equation • always products over reactants • K is called the equilibrium constant • unitless Tro: Chemistry: A Molecular Approach, 2/e

  11. Writing Equilibrium Constant Expressions • For the reaction aA(aq) + bB(aq)ÛcC(aq) + dD(aq) the equilibrium constant expression is • So for the reaction • 2 N2O5(g)Û 4 NO2(g) + O2(g) • the equilibrium constant • expression is Tro: Chemistry: A Molecular Approach, 2/e

  12. What Does the Value of Keq Imply? • When the value of Keq >> 1, we know that when the reaction reaches equilibrium there will be many more product molecules present than reactant molecules • the position of equilibrium favors products • When the value of Keq << 1, we know that when the reaction reaches equilibrium there will be many more reactant molecules present than product molecules • the position of equilibrium favors reactants Tro: Chemistry: A Molecular Approach, 2/e

  13. A Large Equilibrium Constant Tro: Chemistry: A Molecular Approach, 2/e

  14. A Small Equilibrium Constant Tro: Chemistry: A Molecular Approach, 2/e

  15. Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following 2 SO2(g) + O2(g) Û 2 SO3(g) K = 8 x 1025 N2(g) + 2 O2(g) Û 2 NO2(g) K = 3 x 10−17 favors products favors reactants Tro: Chemistry: A Molecular Approach, 2/e

  16. Relationships between Kand Chemical Equations • When the reaction is written backward, the equilibrium constant is inverted For the reaction aA + bB ÛcC + dD the equilibrium constant expression is For the reaction cC + dD ÛaA + bB the equilibrium constant expression is Tro: Chemistry: A Molecular Approach, 2/e

  17. Relationships between Kand Chemical Equations • When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor For the reaction aA + bB ÛcC the equilibrium constant expression is For the reaction 2aA + 2bB Û 2cC the equilibrium constant expression is Tro: Chemistry: A Molecular Approach, 2/e

  18. Relationships between Kand Chemical Equations • When you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations For the reactions (1) aA ÛbB and (2) bBÛcC the equilibrium constant expressions are For the reaction aA ÛcC the equilibrium constant expression is Tro: Chemistry: A Molecular Approach, 2/e

  19. K K’ Example 14.2: Compute the equilibrium constant at 25 °C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g) Given: Find: for N2(g) + 3 H2(g) Û 2 NH3(g), K = 3.7 x 108 at 25 °C K for NH3(g)  0.5N2(g) + 1.5H2(g), at 25 °C Conceptual Plan: Relationships: Kbackward = 1/Kforward, Knew = Koldn Solution: N2(g) + 3 H2(g) Û 2 NH3(g) K1 = 3.7 x 108 2 NH3(g)Û N2(g) + 3 H2(g) NH3(g)Û 0.5 N2(g) + 1.5 H2(g) Tro: Chemistry: A Molecular Approach, 2/e

  20. Practice – When the reaction A(aq)Û 2 B(aq) reaches equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M. When the reaction 2 B(aq)Û Z(aq) reaches equilibrium [B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M. Calculate the equilibrium constant for each of these reactions and the equilibrium constant for the reaction 3 Z(aq)Û 3 A(aq) Krxn 1 = 1.6 x 104 Krxn 2 = 5.0 x 10−1 2 B Û A K3 = 1/Krxn 1 = 6.25 x 10−5 Z Û 2 B K4 = 1/Krxn 2 = 2 Z Û A K3K4 = 1.25 x 10−4 3 Z Û 3 A (K3K4)3 = 2.0 x 10−12 Tro: Chemistry: A Molecular Approach, 2/e

  21. Equilibrium Constants for Reactions Involving Gases • The concentration of a gas in a mixture is proportional to its partial pressure • Therefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gases • For aA(g) + bB(g) ÛcC(g) + dD(g) the equilibrium constant expressions are or Tro: Chemistry: A Molecular Approach, 2/e

  22. Kc and Kp • In calculating Kp, the partial pressures are always in atm • The values of Kp and Kc are not necessarily the same • because of the difference in units • Kp = Kc when Dn = 0 • The relationship between them is Dn is the difference between the number of moles of reactants and moles of products Tro: Chemistry: A Molecular Approach, 2/e

  23. Deriving the Relationshipbetween Kp and Kc Tro: Chemistry: A Molecular Approach, 2/e

  24. Deriving the RelationshipBetween Kp and Kc for aA(g) + bB(g)  cC(g) + dD(g) substituting Tro: Chemistry: A Molecular Approach, 2/e

  25. Kp Kc Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) Û 2 NO2(g), given Kp = 2.2 x 1012 @ 25 °C Given: Find: Kp = 2.2 x 1012 Kc Conceptual Plan: Relationships: Solution: 2 NO(g) + O2(g)  2 NO2(g) Dn = 2  3 = −1 Check: K is a unitless number because there are more moles of reactant than product, Kc should be larger than Kp, and it is Tro: Chemistry: A Molecular Approach, 2/e

  26. Practice – Calculate the value of Kp or Kc for each of the following at 27 °C 2 SO2(g) + O2(g) Û 2 SO3(g) Kc = 8 x 1025 N2(g) + 2 O2(g) Û 2 NO2(g) Kp = 3 x 10−17 Tro: Chemistry: A Molecular Approach, 2/e

  27. Heterogeneous Equilibria • Pure solids and pure liquids are materials whose concentration doesn’t change during the course of a reaction • its amount can change, but the amount of it in solution doesn’t • because it isn’t in solution • Because their concentration doesn’t change, solids and liquids are not included in the equilibrium constant expression • For the reaction aA(s) + bB(aq) ÛcC(l) + dD(aq) the equilibrium constant expression is Tro: Chemistry: A Molecular Approach, 2/e

  28. Heterogeneous Equilibria The amount of C is different, but the amounts of CO and CO2 remain the same. Therefore the amount of C has no effect on the position of equilibrium. Tro: Chemistry: A Molecular Approach, 2/e

  29. Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following HNO2(aq) + H2O(l)Û H3O+(aq) + NO2−(aq)K = 4.6 x 10−4 Ca(NO3)2(aq) + H2SO4(aq)Û CaSO4(s) + 2 HNO3(aq)K = 1 x 104 Tro: Chemistry: A Molecular Approach, 2/e

  30. Calculating Equilibrium Constants from Measured Equilibrium Concentrations • The most direct way of finding the equilibrium constant is to measure the amounts of reactants and products in a mixture at equilibrium • The equilibrium mixture may have different amounts of reactants and products, but the value of the equilibrium constant will always be the same • as long as the temperature is kept constant • the value of the equilibrium constant is independent of the initial amounts of reactants and products Tro: Chemistry: A Molecular Approach, 2/e

  31. Initial and Equilibrium Concentrations forH2(g) + I2(g) Û 2HI(g) @ 445 °C Tro: Chemistry: A Molecular Approach, 2/e

  32. Calculating Equilibrium Concentrations • Stoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know initial concentrations and one equilibrium concentration • Use the change in the concentration of the material that you know to determine the change in the other chemicals in the reaction Tro: Chemistry: A Molecular Approach, 2/e

  33. Calculating Equilibrium Concentrations:An Example 2 A(aq) + B(aq)Û4 C(aq) • Suppose you have a reaction 2 A(aq) + B(aq)Û 4 C(aq)with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50 M. Product C increased by 0.50 mol/L Because Product C increased by 0.50 mol/L, you must have used ½ that amount of Reactant A, according to the stoichiometry. Because Product C increased by 0.50 mol/L, you must have used ¼ that amount of Reactant B, according to the stoichiometry. -½(0.50) -¼(0.50) +0.50 0.88 0.75 Tro: Chemistry: A Molecular Approach, 2/e

  34. Example 14.6:Find the value of Kc for the reaction2 CH4(g) Û C2H2(g) + 3 H2(g) at 1700 °C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M +0.035 Tro: Chemistry: A Molecular Approach, 2/e

  35. Example 14.6:Find the value of Kc for the reaction2 CH4(g) Û C2H2(g) + 3 H2(g) at 1700 °C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M −2(0.035) +3(0.035) +0.035 0.105 0.045 Tro: Chemistry: A Molecular Approach, 2/e

  36. Practice –The following data were collected for the reaction 2 NO2(g) Û N2O4(g) at 100 °C. Complete the table and determine values of Kp and Kc. Tro: Chemistry: A Molecular Approach, 2/e

  37. Practice – A second experiment was done and the following data were collected for the reaction 2 NO2(g) Û N2O4(g) at 100 °C. Complete the table and determine values of Kp and Kc. Compare them to the first experiment. Tro: Chemistry: A Molecular Approach, 2/e

  38. The Reaction Quotient for the gas phase reaction aA + bB ÛcC + dD the reaction quotient is: • If a reaction mixture, containing both reactants and products, is not at equilibrium; how can we determine in which direction it will proceed? • The answer is to compare the current concentration ratios to the equilibrium constant • The concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q Tro: Chemistry: A Molecular Approach, 2/e

  39. The Reaction Quotient:Predicting the Direction of Change • If Q > K, the reaction will proceed fastest in the reverse direction • the [products] will decrease and [reactants] will increase • If Q < K, the reaction will proceed fastest in the forward direction • the [products] will increase and [reactants] will decrease • If Q = K, the reaction is at equilibrium • the [products] and [reactants] will not change • If a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction • If a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction Tro: Chemistry: A Molecular Approach, 2/e

  40. Q, K, and the Direction of Reaction Tro: Chemistry: A Molecular Approach, 2/e

  41. PI2, PCl2, PICl Q Example 14.7: For the reaction below, which direction will it proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm Given: Find: for I2(g) + Cl2(g) Û 2 ICl(g), Kp = 81.9 direction reaction will proceed Conceptual Plan: Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse Solution: I2(g) + Cl2(g) Û 2 ICl(g) Kp = 81.9 because Q (10.8) < K (81.9), the reaction will proceed to the right Tro: Chemistry: A Molecular Approach, 2/e

  42. Practice – For the reaction CH4(g) + 2 H2S(g) Û CS2(g) + 4 H2(g) Kc = 3.59 at 900 °C. For each of the following measured concentrations determine whether the reaction is at equilibrium. If not at equilibrium, in which direction will the reaction proceed to get to equilibrium? Tro: Chemistry: A Molecular Approach, 2/e

  43. K, [COF2], [CF4] [CO2] Example 14.8: If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and Kc = 2.00 @ 1000 °C, find the [CO2]eq for the reaction given. Given: Find: Sort: You’re given the reaction and Kc. You’re also given the [X]eq of all but one of the chemicals 2 COF2Û CO2 + CF4 [COF2]eq = 0.255 M, [CF4]eq = 0.118 M [CO2]eq Conceptual Plan: Relationships: Strategize: You can calculate the missing concentration by using the equilibrium constant expression Solution: Solve: Solve the equilibrium constant expression for the unknown quantity by substituting in the given amounts Check: substitute approximations back in to see if the value agrees Check:  Tro: Chemistry: A Molecular Approach, 2/e

  44. Practice – A sample of PCl5(g) is placed in a 0.500 L container and heated to 160 °C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3 and Cl2 are formed. Determine the equilibrium concentration of PCl5 if Kc = 0.0635 PCl5Û PCl3 + Cl2 Tro: Chemistry: A Molecular Approach, 2/e

  45. Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressures Step 1: decide which direction the reaction will proceed • compare Q to K Step 2: define the changes of all materials in terms of x • use the coefficient from the chemical equation as the coefficient of x • the x change is + for materials on the side the reaction is proceeding toward • the x change is  for materials on the side the reaction is proceeding away from Step 3: solve for x • for 2nd order equations, take square roots of both sides or use the quadratic formula • may be able to simplify and approximate answer for very large or small equilibrium constants Tro: Chemistry: A Molecular Approach, 2/e

  46. Example 14.11:For the reaction I2(g) + Cl2(g) Û 2 ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations Qp(1) < Kp(81.9), so the reaction is proceeding forward Tro: Chemistry: A Molecular Approach, 2/e

  47. Example 14.11:For the reaction I2(g) + Cl2(g) Û 2 ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations +2x x x 0.100+2x 0.100x 0.100x Tro: Chemistry: A Molecular Approach, 2/e

  48. Example 14.11:For the reaction I2(g) + Cl2(g) Û 2 ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations x x +2x 0.100x 0.100x 0.100+2x Tro: Chemistry: A Molecular Approach, 2/e

  49. Example 14.11:For the reaction I2(g) + Cl2(g) Û 2 ICl(g) @ 25°C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations +2x x x −0.0729 −0.0729 +2(0.0729) 0.100+2x 0.100x 0.027 0.100x 0.027 0.246 Tro: Chemistry: A Molecular Approach, 2/e

  50. Example 14.11:For the reaction I2(g) + Cl2(g) Û 2 ICl(g) @ 25 °C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations −0.0729 −0.0729 2(0.0729) 0.027 0.027 0.246 Kp(calculated) = Kp(given) within significant figures Tro: Chemistry: A Molecular Approach, 2/e

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