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Chemical Equilibrium

Chemical Equilibrium. The reversibility of reactions. Equilibrium. Many chemical reactions do not go to completion. Initially, when reactants are present, the forward reaction predominates. As the concentration of products increases, the reverse reaction begins to become significant.

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Chemical Equilibrium

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  1. Chemical Equilibrium The reversibility of reactions

  2. Equilibrium Many chemical reactions do not go to completion. Initially, when reactants are present, the forward reaction predominates. As the concentration of products increases, the reverse reaction begins to become significant.

  3. Equilibrium The forward reaction rate slows down since the concentration of reactants has decreased. Since the concentration of products is significant, the reverse reaction rate gets faster. Eventually, the forward reaction rate = reverse reaction rate

  4. Equilibrium forward reaction rate = reverse reaction rate At this point, the reaction is in equilibrium. The equilibrium is dynamic. Both the forward and reverse reactions occur, but there is no net change in concentration.

  5. Equilibrium: 2NO2↔ N2O4 The concentrations of N2O4 and NO2 level off when the system reaches equilibrium.

  6. Equilibrium: 2NO2↔ N2O4 The system will reach equilibrium starting with either NO2 , N2O4, or a mixture of the two.

  7. Equilibrium Reactions that can reach equilibrium are indicated by double arrows (↔ or ). The extent to which a reaction proceeds in a particular direction depends upon the reaction, temperature, initial concentrations, pressure (if gases), etc.

  8. Equilibrium Scientists studying many reactions atequilibrium determined that for a general reaction with coefficients a, b, c and d : a A + b B ↔ c C + d D K is the equilibrium constant for the reaction. [C]c[D]d K= [A]a[B]b

  9. Equilibrium a A + b B ↔ c C + d D This is the equilibrium constant expression for the reaction. K is the equilibrium constant. The square brackets indicate the concentrations of products and reactants at equilibrium. [C]c[D]d K= [A]a[B]b

  10. Equilibrium Constants The value of the equilibrium constant depends upon temperature, but does not depend upon the initial concentrations of reactants and products. It is determined experimentally.

  11. Equilibrium Constants

  12. Equilibrium Constants The units of K are usually omitted. The square brackets in the equilibrium constant expression indicate concentration in moles/liter. Some texts will use the symbol Kc or Keq for equilibrium constants that use concentrations or molarity.

  13. Equilibrium Constants The size of the equilibrium constant gives an indication of whether a reaction proceeds to the right.

  14. Equilibrium Constants If a reaction has a small value of K, the reverse reaction will have a large value of K. At a given temperature, K = 1.3 x 10-2 for: N2(g) + 3 H2(g) ↔ 2 NH3(g) If the reaction is reversed, 2 NH3(g) ↔ N2(g) + 3 H2(g) K = 1/(1.3 x 10-2 ) = 7.7 x 101

  15. Equilibrium Constants Equilibrium constants for gas phase reactions are sometimes determined using pressures rather than concentrations. The symbol used is Kp rather than K (or Kc). For the reaction: N2(g) + 3 H2(g) ↔ 2 NH3(g) Kp = (Pammonia)2 (PN2)(PH2)3

  16. Equilibrium Constants The numerical value of Kp is usually different than that for K. The values are related, since P= nRT = MRT, where M = mol/liter. The method for solving equilibrium problems with Kp or K is the same. With Kp, you use pressure in atmospheres. With K or Kc, you use concentration in moles/liter. V

  17. Heterogeneous Equilibria Many equilibrium reactions involve reactants and products where more than one phase is present. These are called heterogeneous equilibria. An example is the equilibrium between solid calcium carbonate and calcium oxide and carbon dioxide. CaCO3(s) ↔ CaO(s) + CO2(g)

  18. Heterogeneous Equilibria CaCO3(s) ↔ CaO(s) + CO2(g) Experiments show that the position of the equilibrium does not depend upon the amounts of calcium carbonate or calcium oxide. That is, adding or removing some of the CaCO3(s) or CaO(s) does not disrupt or alter the concentration of carbon dioxide.

  19. Heterogeneous Equilibria CaCO3(s) ↔ CaO(s) + CO2(g)

  20. Heterogeneous Equilibria CaCO3(s) ↔ CaO(s) + CO2(g) This is because the concentrations of pure solids (or liquids) cannot change. As a result, the equilibrium constant expression for the above reaction is: K=[CO2] or Kp = PCO2

  21. Heterogeneous Equilibria The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.

  22. Problem: Calculation of K • At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction: N2(g) + 3 H2(g) ↔ 2 NH3(g)

  23. Problem: Calculation of K • At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction: N2(g) + 3 H2(g) ↔ 2 NH3(g) 1. Write the equilibrium constant expression.

  24. Problem: Calculation of K • At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction: N2(g) + 3 H2(g) ↔ 2 NH3(g) 1. Write the equilibrium constant expression. K = [NH3]2/([N2][H2]3)

  25. Problem: Calculation of K • At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction: N2(g) + 3 H2(g) ↔ 2 NH3(g) 2. Make a table of concentrations.

  26. Problem: Calculation of K N2(g) + 3 H2(g) ↔ 2NH3(g)

  27. Problem: Calculation of K N2(g) + 3 H2(g) ↔ 2NH3(g) 3. Complete the table.

  28. Problem: Calculation of K N2(g) + 3 H2(g) ↔ 2NH3(g) 3. Complete the table.

  29. Problem: Calculation of K N2(g) + 3 H2(g) ↔ 2NH3(g) 3. Complete the table.

  30. N2(g) + 3H2(g) ↔2NH3(g)

  31. N2(g) + 3H2(g) ↔2NH3(g) The equilibrium values are the sum of the initial concentrations plus any changes that occur.

  32. N2(g) + 3H2(g) ↔2NH3(g) The equilibrium values are the sum of the initial concentrations plus any changes that occur.

  33. N2(g) + 3H2(g) ↔2NH3(g) 4. Substitute and solve for K. K = [NH3]2/([N2][H2]3)

  34. N2(g) + 3H2(g) ↔2NH3(g) 4. Substitute and solve for K. K = [NH3]2/([N2][H2]3) K=(1.25)2/[(.13)(.38)3] = 2.2 x 102

  35. The Reaction Quotient If reactants and products of a reaction are mixed together, it is possible to determine whether the reaction will proceed to the right or left. This is accomplished by comparing the composition of the initial mixture to that at equilibrium.

  36. The Reaction Quotient If the concentration of one of the reactants or products is zero, the reaction will proceed so as to make the missing component. If all components are present initially, the reaction quotient, Q, is compared to K to determine which way the reaction will go.

  37. The Reaction Quotient Q has the same form as the equilibrium constant expression, but we use initial concentrations instead of equilibrium concentrations. Initial concentrations are usually indicated with a subscript zero.

  38. The Reaction Quotient In the previous problem, we determined that K = 2.2 x 102 for the reaction: N2(g) + 3 H2(g) ↔ 2 NH3(g) • Suppose 1.00 mol N2, 2.0 mol H2 and 1.5 mol of NH3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed?

  39. The Reaction Quotient K = 2.2 x 102 for the reaction: N2(g) + 3 H2(g) ↔ 2 NH3(g) • Suppose 1.00 mol N2, 2.0 mol H2 and 1.5 mol of NH3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed? 1. Calculate Q and compare it to K.

  40. The Reaction Quotient A comparison of Q with K will indicate the direction the reaction will go.

  41. The Reaction Quotient K = 2.2 x 102 for the reaction: N2(g) + 3 H2(g) ↔ 2 NH3(g) • Suppose 1.00 mol N2, 2.0 mol H2 and 1.5 mol of NH3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed? 1. Calculate Q and compare it to K.

  42. Problem • H2(g) + I2(g) ↔ 2HI(g) Kp = 1.00 x 102 Initially, Phydrogen = Piodine = 0.500 atm. Calculate the equilibrium partial pressures of all three gases.

  43. Problem • H2(g) + I2(g) ↔ 2HI(g) Kp = 1.00 x 102 Initially, Phydrogen = Piodine = 0.500 atm. Calculate the equilibrium partial pressures of all three gases. 1. Write the Kpexpression.

  44. Problem • H2(g) + I2(g) ↔ 2HI(g) Kp = 1.00 x 102 Initially, Phydrogen = Piodine = 0.500 atm. Calculate the equilibrium partial pressures of all three gases. 1. Write the Kpexpression. Kp = (PHI)2 (PH2)(PI2)

  45. Problem • H2(g) + I2(g) ↔ 2HI(g) Kp = 1.00 x 102 Initially, Phydrogen = Piodine = 0.500 atm. Calculate the equilibrium partial pressures of all three gases. 2. Make a table of initial, change and equilibrium pressures.

  46. Problem: H2(g) + I2(g) ↔ 2HI(g)

  47. Problem: H2(g) + I2(g) ↔ 2HI(g)

  48. Problem: H2(g) + I2(g) ↔ 2HI(g)

  49. Problem: H2(g) + I2(g) ↔ 2HI(g) 3. Substitute and solve.

  50. Problem: H2(g) + I2(g) ↔ 2HI(g) • Substitute and solve. • Kp= (PHI)2/(PH2)(PI2) = 1.00 x 102 • (2x)2= 1.00 x 102 • (.500-x) (.500-x)

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