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Chapter 13

Chapter 13 . Chemical Equilibrium. N 2 O 4 ( g ). 2NO 2 ( g ). The Equilibrium State. Chemical Equilibrium : The state reached when the concentrations of reactants and products remain constant over time. Brown. Colorless. N 2 O 4 ( g ). 2NO 2 ( g ). The Equilibrium State.

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Chapter 13

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  1. Chapter 13 Chemical Equilibrium

  2. N2O4(g) 2NO2(g) The Equilibrium State Chemical Equilibrium: The state reached when the concentrations of reactants and products remain constant over time. Brown Colorless

  3. N2O4(g) 2NO2(g) The Equilibrium State

  4. The Equilibrium Constant Kc

  5. aA + bB cC + dD [C]c[D]d [A]a[B]b The Equilibrium Constant Kc For a general reversible reaction: Double arrows show reversible reaction Products Equilibrium equation: Kc = Reactants Equilibrium constant Equilibrium constant expression

  6. N2O4(g) 2NO2(g) [NO2]2 [N2O4] The Equilibrium Constant Kc For the following reaction: Kc = = 4.64 x 10-3 (at 25 °C)

  7. (0.0125)2 (0.0141)2 [NO2]2 0.0337 0.0429 [N2O4] Learning Check Experiment 2 Experiment 5 = 4.63 x 10-3 Kc = = 4.64 x 10-3 =

  8. [N2][H2]3 [NH3]2 [NH3]4 [N2]2[H2]6 [N2][H2]3 [NH3]2 2N2(g) + 6H2(g) N2(g) + 3H2(g) 2NH3(g) 2NH3(g) 4NH3(g) N2(g) + 3H2(g) 1 Kc The Equilibrium Constant Kc The equilibrium constant and the equilibrium constant expression are for the chemical equation as written. Kc = ´ Kc = = ´´ 2 K c = = Kc

  9. Examples • Write the equilibrium constant, Kc and K’c for the following reactions: CH4(g) + H2O(g) CO(g) + 3 H2(g)

  10. 2 P NO2 N2O4(g) 2NO2(g) P N2O4 The Equilibrium Constant Kp Kp = P is the partial pressure of that component

  11. L atm 0.082058 K mol The Equilibrium Constant Kp Kp = Kc(RT)Dn R is the gas constant, T is the absolute temperature (Kelvin). n is the number of moles of gaseous products minus the number of moles of gaseous reactants.

  12. [CaO][CO2] (1)[CO2] [CaCO3] CaCO3(s) CaO(s) + CO2(g) (1) Kp = P CO2 Heterogeneous Equilibria Limestone Lime Kc = = = [CO2] Pure solids and pure liquids are not included. Kc = [CO2]

  13. Examples • In the industrial synthesis of hydrogen, mixtures of CO and H2 are enriched in H2 by allowing the CO to react with steam. The chemical equation for this so-called water-gas shift reaction is CO(g) + H2O(g) CO2(g) + H2(g) What is the value of Kp at 700K if the partial pressures in an equilibrium mixture at 700K are 1.31 atm of CO, 10.0 atm of H2O, 6.12 atm of CO2, and 20.3 atm of H2?

  14. Examples • Consider the following unbalanced reaction (NH4)2S(s) 2NH3(g) + H2S(g) An equilibrium mixture of this mixture at a certain temperature was found to have [NH3] = 0.278 M and [H2S] = 0.355 M. What is the value of the equilibrium constant (Kc) at this temperature?

  15. Examples • Consider the following reaction and corresponding value of Kc: H2(g) + I2(s) 2HI(g) Kc = 6.2 x 102 at 25oC What is the value of Kp at this temperature?

  16. Heterogeneous Equilibria

  17. 2H2(g) + O2(g) H2(g) + I2(g) 2H2O(g) 2H2(g) + O2(g) 2H2O(g) 2HI(g) Using the Equilibrium Constant (at 500 K) Kc = 4.2 x 10-48 (at 500 K) Kc = 57.0 Kc = 2.4 x 1047 (at 500 K)

  18. [C]tc[D]td [A]ta[B]tb aA + bB cC + dD Using the Equilibrium Constant Reaction quotient: Qc= The reaction quotient, Qc, is defined in the same way as the equilibrium constant, Kc, except that the concentrations in Qcare not necessarily equilibrium values.

  19. Using the Equilibrium Constant net reaction goes from left to right (reactants to products). • If Qc < Kc net reaction goes from right to left (products to reactants). • If Qc > Kc no net reaction occurs. • If Qc = Kc

  20. Finding Equilibrium concentrations from Initial Concentrations • Steps to follow in calculating equilibirium concentrations from initial concentration • Write a balance equation for the reaction • Make an ICE (Initial, Change, Equilibrium) table, involves • The initial concentrations • The change in concentration on going to equilibrium, defined as x • The equilibrium concentration • Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x • Calculate the equilibrium concentrations form the calculated value of x • Check your answers

  21. Using the Equilibrium Constant • Sometimes you must use quadratic equation to solve for x, choose the mathematical solution that makes chemical sense Quadratic equation ax2 + bx + c = 0

  22. Finding Equilibrium concentrations from Initial Concentrations [products] For Homogeneous Mixture Kc = [reactants] aA(g) bB(g) + cC(g) Initial concentration (M) [A]initial [B]initial [C]initial Change (M) - ax + bx + cx Equilibrium (M) [A]initial – ax [B]initial – bx [C]initial - cx

  23. Finding Equilibrium concentrations from Initial Concentrations [products] Kc = For Heterogenous Mixture [reactants] aA(g) bB(g) + cC(s) Initial concentration (M) [A]initial [B]initial …….. Change (M) - ax + bx …….. Equilibrium (M) [A]initial – ax [B]initial + bx ……….

  24. H2(g) + I2(g) 2HI(g) Using the Equilibrium Constant At 700 K, 0.500 mol of HI is added to a 2.00 L container and allowed to come to equilibrium. Calculate the equilibrium concentrations of H2, I2, and HI . Kc is 57.0 at 700 K.

  25. Examples • Consider the following reaction I2(g) + Cl2(g) 2ICl(g) Kp = 81.9 (at 25oC) A reaction mixture at 25oC intially contains PI2 = 0.100 atm, PCl2 = 0.100 atm, and PICl = 0.100 atm. Find the equilibrium pressure of I2, Cl2 and ICl at this temperature. In which direction does the reaction favored?

  26. Examples • The value of Kc for the reaction is 3.0 x 10-2. Determine the equilibrium concentration if the initial concentration of water is 8.75 M C(s) + H2O(g) CO(g) + H2(g)

  27. Le Châtelier’s Principle Le Châtelier’s Principle: If a stress is applied to a reaction mixture at equilibrium, net reaction occurs in the direction that relieves the stress. • The concentration of reactants or products can be changed. • The pressure and volume can be changed. • The temperature can be changed.

  28. N2(g) + 3H2(g) 2NH3(g) Altering an Equilibrium Mixture: Concentration

  29. Altering an Equilibrium Mixture: Concentration

  30. Altering an Equilibrium Mixture: Concentration In general, when an equilibrium is disturbed by the addition or removal of any reactant or product, Le Châtelier’s principle predicts that • the concentration stress of an added reactant or product is relieved by net reaction in the direction that consumesthe added substance. • the concentration stress of a removedreactant or product is relieved by net reaction in the direction that replenishes the removed substance.

  31. Altering an Equilibrium Mixture: Concentration • Add reactant – denominator in Qc expression becomes larger • Qc < Kc • To return to equilibrium, Qc must be increases • More product must be made => reaction shifts to the right • Remove reactant – denominator in Qc expression becomes smaller • Qc > Kc • To return to equilibrium, Qc must be decreases • Less product must be made => reaction shifts to the left

  32. Altering an Equilibrium Mixture: Concentration

  33. [NH3]2 [N2][H2]3 N2(g) + 3H2(g) 2NH3(g) (1.98)2 (1.50)(3.00)3 Altering an Equilibrium Mixture: Concentration at 700 K, Kc = 0.291 An equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH3 is disturbed by increasing the N2 concentration to 1.50 M. Qc = = = 0.0968 < Kc Since Qc < Kc, more reactants will be consumed and the net reaction will be from left to right.

  34. Examples • The reaction of iron (III) oxide with carbon monoxide occurs in a blast furnace when iron ore is reduced to iron metal: Fe2O3(s) + 3 CO(g) 2 Fe(l) + 3CO2(g) Use Le Chatellier’s principle to predict the direction of net reaction when an equilibrium mixture is disturbed by: a. adding Fe2O3 b. Removing CO2 c. Removing CO; also account for the change using the reaction quotient Qc

  35. Examples • Consider the following reaction at equilibrium CO(g) + Cl2(g) COCl2(g) Predict whether the reaction will shift left, shift right, or remain unchanged upon each of the following reaction mixture a. COCl2 is added to the reaction mixture b. Cl2 is added to the reaction mixture c. COCl2 is removed from the reaction mixture: also account for the change using the reaction quotient Qc

  36. N2(g) + 3H2(g) 2NH3(g) Altering an Equilibrium Mixture: Pressure and Volume

  37. Altering an Equilibrium Mixture: Pressure and Volume In general, when an equilibrium is disturbed by a change in volume which results in a corresponding change in pressure, Le Châtelier’s principle predicts that • an increasein pressure by reducing the volume will bring about net reaction in the direction that decreases the number of moles of gas. • a decreasein pressure by enlarging the volume will bring about net reaction in the direction that increasesthe number of moles of gas.

  38. Altering an Equilibrium Mixture: Pressure and Volume • If reactant side has more moles of gas • Denominator will be larger • Qc < Kc • To return to equilibrium, Qc must be increased • Reaction shifts toward fewer moles of gas (to the product) • If product side has more moles of gas • Numerator will be larger • Qc > Kc • To return to equilibrium, Qc must be decreases • Reaction shifts toward fewer moles of gas (to the reactant)

  39. Altering an Equilibrium Mixture: Pressure and Volume • Reaction involves no change in the number moles of gas • No effect on composition of equilibrium mixture • For heterogenous equilibrium mixture • Effect of pressure changes on solids and liquids can be ignored • Volume is nearly independent of pressure • Change in pressure due to addition of inert gas • No change in the molar concentration of reactants or products • No effect on composition

  40. Examples • Consider the following reaction at chemical equilibrium 2 KClO3(s) 2 KCl(s) + 3O2(g) a. What is the effect of decreasing the volume of the reaction mixture? b. Increasing the volume of the reaction mixture? c. Adding inert gas at constant volume?

  41. Examples • Does the number moles of products increases, decreases or remain the same when each of the following equilibria is subjected to a increase in pressure by decreasing the volume? • PCl5(g) PCl3(g) + Cl2(g) • CaO(s) + CO2(g) CaCO3(s) • 3 Fe(s) + 4H2O(g) Fe3O4(s) + 4 H2(g)

  42. N2(g) + 3H2(g) 2NH3(g) Altering an Equilibrium Mixture: Temperature DH° = -2043 kJ (exothermic ) As the temperature increases, the equilibrium shifts from products to reactants.

  43. Altering an Equilibrium Mixture: Temperature

  44. Altering an Equilibrium Mixture: Temperature In general, when an equilibrium is disturbed by a change in temperature, Le Châtelier’s principle predicts that • the equilibrium constant for an exothermicreaction (negative DH°) decreasesas the temperature increases. • Contains more reactant than product • Kc decreases with increasing temperature • the equilibrium constant for an endothermicreaction (positive DH°) increasesas the temperature increases. • Contains more product than reactant • Kc increases with increasing temperature

  45. Examples • The following reaction is endothermic CaCO3(s) CaO(s) + CO2(g) a. What is the effect of increasing the temperature of the reaction mixture? b. Decreasing the temperature?

  46. Examples • In the first step of Ostwald process for the synthesis of nitric acid, ammonia is oxidized to nitric oxide by the reaction: 2 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(l) ΔHo = -905 kJ How does the temperature amount of NO vary with an increases in temperature?

  47. The Effect of a Catalyst on Equilibrium

  48. The Effect of a Catalyst on Equilibrium • Catalyst increases the rate of a chemical reaction • Provide a new, lower energy pathway • Forward and reverse reactions pass through the same transition state • Rate for forward and reverse reactions increase by the same factor • Does not affect the composition of the equilibrium mixture • Does not appear in the balance chemical equation • Can influence choice of optimum condition for a reaction

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