1. Chapter 18 Chemical Equilibrium 18.1 The Nature of Chemical Equilibrium

2. Reversible Reactions A chemical reaction in which the products re-form the reactants Reactants ↔ Products Le Chatelier’s Principle When system at equilibrium is disturbed by application of a stress, it attains a new equilibrium position that minimizes the stress.

3. Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time All reactions carried out in a closed vessel will reach equilibrium

4. Chemical Equilibrium Sometimes, if little product is formed, equilibrium lies far to the left (towards reactants) Reactants Products Sometimes, if little reactant remains, equilibrium lies far to the right (towards products) ReactantsProducts

5. Dynamic Equilibrium Reactions continue to take place Reactant molecules continue to be converted to product Product continues to be converted to reactant (reverse reaction) Forward and reverse reactions take place at the same rate at equilibrium

6. Causes of Equilibrium Beginning of Reaction Only reactant molecules exist, so only reactant molecules may collide Middle As product concentration increases, collisions may take place that lead to the reverse reaction At Equilibrium Rates of forward and reverse reactions are identical

7. Suppose that compound A reacts with B to form C and D. The symbol is used for system at equilibrium.

8. If we add more A what would happen to the amount of C at equilibrium? Adding A increases the amount of C and D at equilibrium.

9. If we add more C what would happen to the amount of A at equilibrium? adding more C increases the amount of A and B at equilibrium.

10. If we add A what would happen to the amount of B at equilibrium? A reacts with B to form C and D. Therefore, adding A decreases B.

11. Example - The Haber Process N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) Hydrogen is consumed at 3x the rate of nitrogen Ammonia is formed at 2x the rate at which nitrogen is consumed

12. Points to Remember 1.Rxns in a closed system always reach equilibrium 2.When a rxn has reached equilibrium, that means the rate of the forward rxn equals the rate of the reverse rxn 3.A system will remain at equilibrium until the system is disturbed in some manner

13. The Equilibrium Constant For the balanced forward equation: jA + kB → lC + mD (j, k, l, m) are coefficients K is a constant called the equilibrium constant K varies depending on temperature and upon the coefficients of the balanced equation K is determined experimentally [X] represents concentration of chemical species at equilibrium K = [C] l [D] m [A] j [B] k

14. Equilibrium Constant 4NH 3(g) + 7O 2(g)  4NO 2(g) +6H 2 O (g) K = [NO 2 ] 4 [H 2 O] 6 [NH 3 ] 4 [O 2 ] 7 The value of K will tell us the position of equilibrium

15. Practice N 2(g) + 3H 2(g)  2NH 3(g) K = [NH 3 ] 2 [N 2 ] [H 2 ] 3 [NH 3 ] = 3.1 x 10 -2 M [N 2 ] = 8.5 x 10 -1 M [H 2 ] = 3.1 x 10 -3 M Calculate K K = [3.1x 10 -2 ] 2 [8.5 x 10 -1 ] [3.1 x 10 -3 ] 3 K = 3.8 x 10 4 no units [equilibrium]

16. Practice N 2(g) + 3H 2(g)  2NH 3(g) [NH 3 ] = 3.1 x 10 -2 M [N 2 ] = 8.5 x 10 -1 M [H 2 ] = 3.1 x 10 -3 M K = 3.8 x 10 4 The value of K will tell us the position of equilibrium If K >1, the rxn favors products If K <1, the rxn favors reactants If K = 1, equilibrium lies in the middle

17. Practice N 2(g) + 3H 2(g)  2NH 3(g) [NH 3 ] = 3.1 x 10 -2 M [N 2 ] = 8.5 x 10 -1 M [H 2 ] = 3.1 x 10 -3 M Calculate new K K = 3.8 x 10 4 2NH 3(g)  N 2(g) + 3H 2(g) Original rxn New rxn Same [ ] Original K The rxn was inversed so K is inversed K = 1 3.8 x 10 4 K = 2.6 x 10 -5

18. Practice N 2(g) + 3H 2(g)  2NH 3 (g) [NH 3 ] = 3.1 x 10 -2 M [N 2 ] = 8.5 x 10 -1 M [H 2 ] = 3.1 x 10 -3 M Calculate new K K = 3.8 x 10 4 Original rxn New rxn Same [ ]Original K The coefficients were reduced by ½ K = (3.8 x 10 4 ) 1/2 K = 2.0 x 10 2 ½N 2(g) + 3/2H 2(g)  NH 3 (g)

19. Interpreting K If K = 1, then there are equal concentrations of reactants & products If K is small, the forward rxn occurs slightly before equilibrium is established, reactants are favored If K is large, reactants are mostly converted into products when equilibrium is established, products are favored

20. Heterogeneous Equilibria The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present If pure solids or liquids are involved in a chemical reaction, their concentrations are not included in the equilibrium expression for the reaction Pure liquids are not the same as solutions, whose concentration can change

21. More Practice! 1.At 25 °C, an equilibrium mixture of gases contains 6.4 x 10 -3 M PCl 3, 2.5 x 10 -2 M Cl 2, & 4.0 x 10 -3 M PCl 5. What is the equilibrium constant for the following reaction? PCl 5 (g) ↔ PCl 3 (g) + Cl 2 (g) 2. At equilibrium, a 2.0 L vessel contains 0.36 mol of H 2, 0.11 mol of Br 2, & 37 mol of HBr. What is the equilibrium constant for the reaction at this temperature? H 2 (g) + Br 2 (g) ↔ HBr (g) K = 4.0 x 10 -2 K = 3.5 x 10 4

22. Chapter 18 Chemical Equilibrium 18.2 Shifting Equilibrium

23. The Effect of Change in Pressure: Ways to Change Pressure 1.Add or remove a gaseous reactant or product 2.Add an inert gas (one not involved in the reaction) An inert gas increases the total pressure but has no effect on the concentrations or partial pressures of the reactants or products

24. The Effect of Change in Pressure 3.Change the volume of the container When the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume. This is done by decreasing the total number of gaseous molecules in the system N 2 (g) + 3H 2 (g) → 2NH 3 (g) shifts to the right to decrease the total molecules of gas present

25. The Effect of Change in Pressure When the container volume is increased, the system will shift so as to increase its volume N 2 (g) + 3H 2 (g) ← 2NH 3 (g) shifts to the left to increase the total number of molecules of gas present

26. The Effect of Change in Concentration If a reactant or product is added to a system at equilibrium, the system will shift away from the added component (it will attempt to "use up" the added component) If a reactant or product is removed from a system at equilibrium, the system will shift toward the removed component (it will attempt to "replace" the removed component)

27. The Effect of a Change in Temperature An increase in temperature increases the energy of the system. Le Chatelier's principle predicts that the system will shift in the direction that consumes the energy For an exothermic rxn, energy is a product. The rxn will shift to the left to use up the excess energy For an endothermic rxn, energy is a reactant. The rxn will shift to the right to use up the energy A decrease in temperature will cause a system shift in the direction that "replaces" the lost energy

28. Le Chatelier’s Principle When a stress is applied to a system at equilibrium, the system will shift in a direction that will relieve the stress [changes] do not affect the value of K pressure changes do not affect the value of K temperature changes will affect the value of K

29. Practice ([change]) As 4 O 6(s) + 6C (s)  As 4(g) + 6CO (g) Which direction will the rxn shift to re-establish equilibrium if: Add CO?leftrightno shift Add or remove C or As 4 O 6 ? leftrightno shift Remove As 4 ? leftrightno shift

30. Practice (change in pressure) Which direction will the rxn shift to reestablish equilibrium if the volume is reduced? P 4(s) + 6Cl 2(g)  4PCl 3(l) leftrightno shift PCl 3(g) + Cl 2(g)  PCl 5(g) leftrightno shift O 2(g) + N 2(g)  2NO (g) leftrightno shift

31. Practice (change in temperature) Which direction will the rxn shift to reestablish equilibrium if the temperature is increased? N 2(g) + O 2(g)  2NO (g) ΔH o = 181kJleftrightno shift 2SO 2(g) + O 2(g)  2SO 3(g) ΔH o = -198kJleftrightno shift How will the change affect K? N 2(g) + O 2(g)  2NO (g) ΔH o = 181kJ 2SO 2(g) + O 2(g)  2SO 3(g) ΔH o = -198kJ increasedecrease increasedecrease

32. Completed Reactions You do not need to consider equilibrium if you have a reaction that goes to completion. Examples: Formation of a gas in an open container Formation of a precipitate Formation of a slightly ionized product (polar) – neutralization

33. The Common-Ion Effect When the addition of an ion common to two solutes brings about precipitation or reduced ionization.

34. Solubility Product and the Common Ion Effect AgCl (s)  Ag + (aq) + Cl - (aq) K sp = [Cl - ][Ag + ] = 1.8 x 10 -10 solublility product What happens if a solution of NaCl is added to this system? NaCl (s)  Na + (aq) + Cl - (aq) rxn shifts  AgCl (s)  Ag + (aq) + Cl - (aq) The solubility of AgCl has actually decreased

35. Chapter 18 Chemical Equilibrium 18.3 Equilibria of Acids, Bases and Salts

36. Ionization Constant of a Weak Acid Since weak acids only partially ionize in solution, an equilibrium exists between the weak acid + water, reactants and the hydronium + conjugate base ions, products. K a = acid ionization constant K a, like K eq, is temperature-dependant

37. KaKa For a weak acid: HF (g) + H 2 O (l) ↔ H 3 O + (aq) + F - (aq) K a = [H 3 O + ] [F - ] [HF] Assume [H 2 O] is constant

38. Ionization Constant of Water Flashback to chapter 15… The self-ionization of water: H 2 O (l) + H 2 O (l) ↔ H 3 O + (aq) + OH - (aq) K W 1.0 x 10 -14 K W = [H 3 O + ][OH - ] = 1.0 x 10 -14 Assume [H 2 O] is constant

39. Hydrolysis of Salts Recall that the conjugates of strong acids & bases are weak, and the conjugates of weak acids & bases are strong: neutralThe salt produced from a strong acid-strong base would be neutral. acidicThe salt produced from a strong acid-weak base would be acidic. basicThe salt produced from a weak acid-strong base would be basic.

40. Hydrolysis A reaction between water molecules and ions of a dissolved salt. If the anions react with water, the process is called anion hydrolysis, and produces a basic pH If the cations react with water, the process is called cation hydrolysis, and produces an acidic pH

41. Anion Hydrolysis For a weak acid: HA (g) + H 2 O (l) ↔ H 3 O + (aq) + A - (aq) K a = [H 3 O + ] [A - ] [HA] The A - can further react with water: A - (aq) + H 2 O (l) ↔ HA (aq) + OH - (aq) (HOH)

42. Anion Hydrolysis A - (aq) + H 2 O (l) ↔ HA (aq) + OH - (aq) The forward reaction is dependant on the strength of the A -, so if the K a of HA is low, the strength of A - will be high.

43. Cation Hydrolysis For a weak base: B (aq) + H 2 O (l) ↔ BH + (aq) + OH - (aq) K b = [BH + ] [OH - ] [B] The BH + can further react with water: BH + (aq) + H 2 O (l) ↔ H 3 O + (aq) + B (aq)

44. Cation Hydrolysis BH + (aq) + H 2 O (l) ↔ H 3 O + (aq) + B (aq) The forward reaction is dependant on the strength of the BH +, so if the K b of B is low, the strength of BH + will be high.

45. Hydrolysis in Acid-Base Reactions Since the product of a neutralization is a salt & water, depending on the type of salt formed, the pH of the solution may be acidic, neutral or basic.

46. Strong Acid-Strong Base The dissociated ions from salts produced by these reactions do not undergo hydrolysis, so the pH of these resulting solutions will be neutral.

47. Weak Acid-Strong Base The dissociated anions produced by these reactions will undergo hydrolysis to produce hydroxide ions that will raise the pH, creating a basic solution

48. Strong Acid-Weak Base The dissociated cations produced by these reactions will undergo hydrolysis to produce hydronium ions that will lower the pH, creating an acidic solution

49. Chapter 18 Chemical Equilibrium 18.4 Solubility Equilibrium

50. Solubility Substances are considered soluble if at least 1 g will dissolve per 100 g water. Substances are considered insoluble if less than 0.1 g will dissolve per 100 g water. Substances that fall in-between 0.1-1 g are considered partially soluble.

51. K sp The solubility product constant indicates the respective molar concentrations of dissolved ions at equilibrium in a saturated solution. The smaller the K sp value, the more insoluble the compound. See Appendix Table A-13, page 861

52. Practice If the solubility of magnesium sulfate at 20°C is 33.7g per 100.g of water, Calculate K sp. 2) Calculate solubility in mol/L 33.7 g x 1 g H 2 O x 1000 mL x 1 mol MgSO 4 100. g 1 mL H 2 O 1 L 120.4 gMgSO 4 = 2.80 mol/L 4) Solve K sp expression 1) Write the rxn MgSO 4 (s) + H 2 O (l)  Mg +2 (aq) + SO 4 -2 (aq) 3) [Mg +2 ] = [SO 4 -2 ] [Mg +2 ] = [SO 4 -2 ] = 2.80 mol/L K sp = [Mg +2 ] [SO 4 -2 ] = [2.80] [2.80] = 7.84

53. Practice Calculate the solubility of CaCO 3, in mol/L at 25°C given K sp = 2.8 x 10 -9 2) [Ca +2 ] = [CO 3 -2 ] So, [Ca +2 ] = x and [CO 3 -2 ] = x Then, K sp = x 2 Solubility of calcium carbonate = 5.3 x 10 -5 1) Write the rxn & K sp expression CaCO 3 (s) + H 2 O (l)  Ca +2 (aq) + CO 3 -2 (aq) K sp = [Ca +2 ] [CO 3 -2 ] 3) Solve for x K sp = 2.8 x 10 -9 = x 2 X = √ 2.8 x 10 -9

54. More Practice! 1.What is the value of K sp for tin (II) sulfide, given that its solubility is 5.2 x 10 -12 g / 100. g water? 2. What is the solubility in mol/L of manganese (II) sulfide given that its K sp value is 2.5 x 10 -13 ? 3. Calculate the concentration of Zn +2 in a saturated solution of zinc sulfide given that K sp of zinc sulfide = 1.6 x 10 -24. 1) 1.2 x 10 -25 2) 5.0 x 10 -7 mol/L 3) 1.3 x 10 -12 mol/L

55. Practice 500 mL of 0.010 M Ba(NO 3 ) 2 and 300 mL of 0.020 M NaF solutions are mixed. Will a precipitate of BaF 2 (K sp = 1.7 x 10 -6 ) form? K sp = [Ba 2+ ][F - ] 2 = 1.7 x 10 -6 BaF 2(s)  Ba 2+ (aq) + 2F - (aq) [initial] Q sp = (.00625)(.0075) 2 = 3.52 x 10 -7 Q < K, so the rxn will shift in favor of the products = no precipitate

56. More Practice! 1)Will a precipitate form if 20. mL of 0.034 M sodium chloride and 15 mL of 0.083 M copper (I) nitrate are mixed? 2) Does a precipitate form if 100. mL of 0.0014 M calcium nitrate and 200. mL of 0.00020 M sodium sulfate are mixed? Yes, CuCl precipitates No, CaSO 4 does not precipitate

57. What if you are asked to find an [equilbrium]? Use the R.I.C.E method!

58. Practice 4NH 3(g) + 3O 2(g)  2N 2(g) + 6H 2 O (g) [NH 3 ] o =.0150 M [O 2 ] o =.0150 M [N 2 ] eq = 1.96 x 10 -3 M When 0.0150 mol of NH 3 and 0.0150 mol of O 2 are placed in a 1.00 L container at a certain temperature, the N 2 concentration at equilibrium is 1.96 x 10 -3 M. Calculate K c for the reaction at this temperature..0150.0150  0 0.0150.0150  0 0 - 4x - 3x  + 2x + 6x - 4x - 3x  + 2x + 6x.0150 - 4x.0150 - 3x  1.96 x 10 -3 6x.0150 - 4x.0150 - 3x  1.96 x 10 -3 6x 4NH 3(g) + 3O 2(g)  2N 2g) + 6H 2 O (g)

59. Practice 4NH 3(g) + 3O 2(g)  2N 2(g) + 6H 2 O (g) [NH 3 ] o =.0150 M [O 2 ] o =.0150 M [N 2 ] eq = 1.96 x 10 -3 M When 0.0150 mol of NH 3 and 0.0150 mol of O 2 are placed in a 1.00 L container at a certain temperature, the N 2 concentration at equilibrium is 1.96 x 10 -3 M. Calculate K c for the reaction at this temperature. K = [H 2 O] 6 [N 2 ] 2 [NH 3 ] 4 [O 2 ] 3 K = (6x) 6 (1.96 x 10 -3 ) 2 (.0150 – 4x) 4 (.0150 – 3x) 3.0150 - 4x.0150 - 3x  1.96 x 10 -3 6x.0150 - 4x.0150 - 3x  1.96 x 10 -3 6x 2x =1.96 x 10 -3 x =9.80 x 10 -4 K =5.95 x 10 -6 Substitute x for this value

60. More Practice Mustard gas, used in chemical warfare in WWI, has been found to be an effective agent in the chemotherapy of Hodgkins disease. It is produced by the following reaction: SCl 2(g) + 2C 2 H 4(g)  S(CH 2 CH 2 Cl) 2(g) An empty five-liter tank is filled with 0.258 mol SCl 2 and 0.592 mol C 2 H 4. After equilibrium is established, 0.0349 mol of mustard gas is present. a)Calculate [equilibrium] for each reactant gas. b)Determine the value of K eq for the reaction. b)K eq = 14.3 a)[SCl 2 ] = 0.0446 M [C 4 H 4 ] = 0.104 M