1. Chemical Equilibrium Chapter 14

2. 14-1 The Equilibrium Constant Objectives To write the equilibrium constant expression for any chemical reaction To determine the relationship between the expression for the equilibrium constant and the form of the balanced equation To convert between equilibrium constants in which the concentrations of gases are expressed in mol/L and those in terms of partial pressures expressed in atmospheres

3. Reversible Reactions: A chemical reaction in which the products can react to re-form the reactants Chemical Equilibrium: When the rate of the forward reaction equals the rate of the reverse reaction and the concentration of products and reactants remains unchanged 2HgO(s)  2Hg(l) + O2(g) Arrows going both directions (  ) indicates equilibrium in a chemical equation

4. 2NO 2 (g)  2NO(g) + O 2 (g) Remember this from Chapter 13? Why was it so important to measure reaction rate at the start of the reaction (method of initial rates?)

5. 2NO 2 (g)  2NO(g) + O 2 (g)

6. Law of Mass Action For the reaction For the reaction: Where K is the equilibrium constant, and is unitless jA + kB  lC + mD

7. Product Favored Equilibrium Large values for K signify the reaction is “product favored” When equilibrium is achieved, most reactant has been converted to product

8. Reactant Favored Equilibrium Small values for K signify the reaction is “reactant favored” When equilibrium is achieved, very little reactant has been converted to product

9. Writing an Equilibrium Expression 2NO 2 (g)  2NO(g) + O 2 (g) K = ??? Write the equilibrium expression for the reaction:

10. 14.1 Chemical Equilibrium cont’d What is K eq for the reaction: 2SO 3 (g) ↔ 2SO 2 (g) + O 2 (g) when the equilibrium concentrations of [SO 2 ] = 0.44M, [O 2 ] = 0.22M, and [SO 3 ] = 0.11M? A: K eq = [SO 2 ] 2 [O 2 ]/[SO 3 ] 2 = 3.5 Remember that: temperature influences K eq K eq is determined by experiment

11. 14.1 Chemical Equilibrium cont’d Note that when a chemical reaction has reached equilibrium, the reaction at the molecular level does not stop! Simply the rate of the forward reaction = the rate of the reverse reaction. For A ↔ B: Since k f [A] = k r [B] then k f /k r = [B]/[A] and K eq = k f /k r = [B]/[A]

12. 14.1 Chemical Equilibrium cont’d What happens if we write the reaction in reverse?

13. Conclusions about Equilibrium Expressions  The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse 2NO 2 (g)  2NO(g) + O 2 (g ) 2NO(g) + O 2 (g)  2NO 2 (g)

14. 14.1 Chemical Equilibrium cont’d What happens if we write the chemical equation with a different set of coefficients?

15. Conclusions about Equilibrium Expressions  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. 2NO 2 (g)  2NO(g) + O 2 (g ) NO 2 (g)  NO(g) + ½O 2 (g )

16. 14.1 Chemical Equilibrium cont’d The equilibrium system of hydrogen, nitrogen, and ammonia can be written several ways: 1. N 2 (g) + 3H 2 (g)  2NH 3 (g) 2. ½N 2 (g) + 3/2 H 2 (g)  NH 3 (g) 3. 2NH 3 (g)  N 2 (g) + 3H 2 (g) Q: The K eq for reaction 1 is 0.19 at 532°C. Write the equilibrium constant expression and calculate K eq for reactions 2 and 3. A: K 2 = 0.44 K 3 = 5.3

17. 14.1 Chemical Equilibrium cont’d How do we measure K eq if we are measuring gases in terms of partial pressure rather than as a concentration in mol/liter? It turns out the ideal gas law can be used to relate K p and K c.

18. Equilibrium Expressions Involving Pressure For the gas phase reaction: 3H 2 (g) + N 2 (g)  2NH 3 (g)

19. 14.1 Chemical Equilibrium cont’d Note that R = 0.08206 L ● atm/mol ● K, T is the temperature in K, and Δn is the change in the number of moles of gas Δn = total number of moles of gas on the product side – the total number of moles of gas on the reactant side Note that Δn can be positive, negative or zero!

20. 14.1 Chemical Equilibrium cont’d Consider the equilibrium: PCl 5 (g)  PCl 3 (g) + Cl 2 (g) If the numerical value of K p = 0.74 at 499 K calculate K c. A: K c = 1.8 x 10 -2

21. 14.1 Chemical Equilibrium cont’d K p for the formation of 2 moles of ammonia from nitrogen and hydrogen is 2.8 x 10 -9 at 298 K calculate K c. N 2 (g) + 3H 2 (g)  2NH 3 (g) A: K c = 1.7 x 10 -6

22. 14-2 The Reaction Quotient Objectives To define and evaluate Q, the reaction quotient To compare Q and K eq to determine in which direction a reaction proceeds when a system is not in equilibrium To create a number line with Q and K eq placed appropriately to determine the direction in which a reaction proceeds

23. 14-2 The Reaction Quotient The law of mass action not only describes equilibrium systems, but provides important information about systems not yet at equilibrium The reaction quotient, Q, has the same algebreic form as K eq, but uses current concentrations rather than equilibrium concentrations in the calculations.

24. 14-2 The Reaction Quotient Why is this useful? Because, comparing Q to K eq enables us to predict the direction in which a reaction will proceed to achieve equilibrium

25. The Reaction Quotient For some time, t, when the system is not at equilibrium, the reaction quotient, Q takes the place of K, the equilibrium constant, in the law of mass action. jA + kB  lC + mD

26. Significance of the Reaction Quotient  If Q = K, the system is at equilibrium  If Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved  If Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

27. 14-2 The Reaction Quotient A scientist mixes 0.5 mol of NO 2 with 0.3 mol of N 2 O 4 in a 2.0L flask at 418K. At this temperature, K c is 0.32 for the reaction of 2 moles of NO 2 to form one mole of N 2 O 4. Does a reaction occur to form more NO 2 or more N 2 O 4 or is the system at equilibrium? A: Q = 2.4 which is greater than 0.32 so reaction proceeds to the left and more NO 2 forms.

28. 14-2 The Reaction Quotient A scientist mixes 0.24 mol of NO 2 with 0.08 mol of N 2 O 4 in a 2.0L flask at 418K. Does a reaction occur to form more NO 2 or more N 2 O 4 or is the system at equilibrium? A: Q = 2.8 which is greater than 0.32 so reaction proceeds to the left and more NO 2 forms.

29. 14-3 The Principle of Le Chatelier Objectives To define the principle of Le Chatelier To predict the response of an equilibrium system to changes in conditions by applying the principle of Le Chatelier To determine how changes in temperature influence the equilibrium system

30. LeChatelier’s Principle When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress. Translated: The system undergoes a temporary shift in order to restore equilibrium.

31. 14-3 The Principle of Le Chatelier Significance? The principle of Le Chatelier can be exploited to increase the yield of a reaction If the products of a reaction can be removed from the reaction mixture, the system responds by producing additional products

32. 14-3 The Principle of Le Chatelier Note that when we talk about a shift in equilibrium occurring we are not changing the Keq !! Rather, we are changing the composition to produce additional products or reactants until equilibrium is restored!!

33. LeChatelier Example #1 A closed container of ice and water is at equilibrium. Then, the temperature is raised. Ice + Energy  Water The system temporarily shifts to the _______ to restore equilibrium. right

34. LeChatelier Example #2 A closed container of N 2 O 4 and NO 2 is at equilibrium. NO 2 is added to the container. N 2 O 4 (g) + Energy  2 NO 2 (g) The system temporarily shifts to the _______ to restore equilibrium. left

35. LeChatelier Example #3 A closed container of water and its vapor is at equilibrium. Vapor is removed from the system. water + Energy  vapor The system temporarily shifts to the _______ to restore equilibrium. right

36. LeChatelier Example #4 A closed container of N 2 O 4 and NO 2 is at equilibrium. The pressure is increased. N 2 O 4 (g) + Energy  2 NO 2 (g) The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation. left

37. 14-4 Equilibrium Calculations Objectives To combine experimental data and stoichiometric relationships to calculate equilibrium constants from experimental data To define a systematic approach to chemical equilibria To write an ICE table for equilibrium problems To calculate the equilibrium concentrations of species in a chemical reaction

38. Solving for Equilibrium Concentration Consider this reaction at some temperature: H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) K = 2.0 Assume you start with 8 molecules of H 2 O and 6 molecules of CO. How many molecules of H 2 O, CO, H 2, and CO 2 are present at equilibrium? “ICE” Here, we learn about “ICE” – the most important problem solving technique in the second semester. You will use it for the next 3 chapters!

39. Solving for Equilibrium Concentration H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) K = 2.0 Step #1: We write the law of mass action for the reaction:

40. Solving for Equilibrium Concentration H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) I Initial: Change: Equilibrium: Step #2: We “ICE” the problem, beginning with the Initial concentrations 8600 -x +x 8-x6-x x x

41. Solving for Equilibrium Concentration Equilibrium: 8-x6-xxx Step #3: We plug equilibrium concentrations into our equilibrium expression, and solve for x H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g)

42. Solving for Equilibrium Concentration Step #4: Substitute x into our equilibrium concentrations to find the actual concentrations H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) Equilibrium: 8-x6-xxx 8-4=46-4=244

43. 14-4 Equilibrium Calculations How do we set up “ICE” tables and determine equilibrium concentrations when the initial [product] ≠ 0? First, you need to calculate Q and compare it to K eq to determine the direction the reaction will proceed!

44. 14.4 Chemical Equilibrium cont’d A researcher studies an industrial process by placing 0.03 mol of sulfuryl chloride, a powerful chemical oxidizer, in a 100-L reactor along with 2.0 mol of SO 2 and 1.0 mol of Cl 2 at 173C. At this temperature, K p is 3.0 for: SO 2 Cl 2 (g)  SO 2 (g) + Cl 2 (g) Worked out in class/see TB

45. 14-4 Equilibrium Calculations How do we set up “ICE” tables and determine equilibrium concentrations when the coefficients for reactants and products are not equal to one?

46. 14.4 Chemical Equilibrium cont’d Write the ICE table and the expression for the equilibrium constant needed to solve for the concentrations of the species when 0.010M SO 2 and 0.05M of O 2 and 0.0020M SO 3 are mixed. At 1009K, K c is 2.0 for: 2SO 2 (g) + O 2 (g)  2SO 3 (g) Worked out in class/see TB

47. 14-5 Heterogeneous Equilibria Objectives To explain why the concentration of a solid or a pure liquid is a constant To write equilibrium constant expressions that describe heterogeneous equilibria

48. 14-5 Heterogeneous Equilibria The concentration of a solid or a pure liquid does not change during the course of a reaction Therefore, the concentrations of solids and pure liquids are not included in the expression for the equilibrium constant. In addition, if any of the substances in the chemical equation is no longer present, the system is not at equilibrium and it cannot be described by the law of mass action

49. 14-5 Heterogeneous Equilibria Heterogeneous equilibria are important to study because chemists often strive to make the product in a different phase than the reactants so that the separation of products from reactants is easier

50. Heterogeneous Equilibria Write the equilibrium expression for the reaction: PCl 5 (s)  PCl 3 (l) + Cl 2 (g) Pure solid Pure liquid

51. 14-6 Solubility Equilibria Objectives To define and write the expression for the solubility product To calculate K sp from experimental data To calculate the solubility of slightly soluble salts from K sp

52. 14-6 Solubility Equilibria Numerous chemical reactions result in the formation of a solid product from reactants in solution Ex: AgNO 3 (aq) + HCl (aq) ↔ AgCl (s) + HNO 3 (aq) Overall Ionic Equation: Ag + (aq) + NO 3 - (aq) + H + (aq) + Cl- (aq) ↔ AgCl (s) + H + (aq) + NO 3 - (aq) Net: Ag + (aq) + Cl - (aq) ↔ AgCl (s) Review writing of net ionic equations from Ch 4!

53. 14-6 Solubility Equilibria Net ionic equations are used to describe solubility equilibria By convention, to determine the solubility product K sp, the equations are written as the dissolving of a solid. Ex: AgCl (s)  Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ]

54. K sp Values for Some Salts at 25  C NameFormulaK sp Barium carbonate BaCO 3 2.6 x 10 -9 Barium chromate BaCrO 4 1.2 x 10 -10 Barium sulfate BaSO 4 1.1 x 10 -10 Calcium carbonate CaCO 3 5.0 x 10 -9 Calcium oxalate CaC 2 O 4 2.3 x 10 -9 Calcium sulfate CaSO 4 7.1 x 10 -5 Copper(I) iodide Cu I 1.3 x 10 -12 Copper(II) iodate Cu( I O 3 ) 2 6.9 x 10 -8 Copper(II) sulfide CuS 6.0 x 10 -37 Iron(II) hydroxide Fe(OH) 2 4.9 x 10 -17 Iron(II) sulfide FeS 6.0 x 10 -19 Iron(III) hydroxide Fe(OH) 3 2.6 x 10 -39 Lead(II) bromide PbBr 2 6.6 x 10 -6 Lead(II) chloride PbCl 2 1.2 x 10 -5 Lead(II) iodate Pb( I O 3 ) 2 3.7 x 10 -13 Lead(II) iodide Pb I 2 8.5 x 10 -9 Lead(II) sulfate PbSO 4 1.8 x 10 -8 NameFormulaK sp Lead(II) bromide PbBr 2 6.6 x 10 -6 Lead(II) chloride PbCl 2 1.2 x 10 -5 Lead(II) iodate Pb( I O 3 ) 2 3.7 x 10 -13 Lead(II) iodide Pb I 2 8.5 x 10 -9 Lead(II) sulfate PbSO 4 1.8 x 10 -8 Magnesium carbonate MgCO 3 6.8 x 10 -6 Magnesium hydroxide Mg(OH) 2 5.6 x 10 -12 Silver bromate AgBrO 3 5.3 x 10 -5 Silver bromide AgBr 5.4 x 10 -13 Silver carbonate Ag 2 CO 3 8.5 x 10 -12 Silver chloride AgCl 1.8 x 10 -10 Silver chromate Ag 2 CrO 4 1.1 x 10 -12 Silver iodate Ag I O 3 3.2 x 10 -8 Silver iodide Ag I 8.5 x 10 -17 Strontium carbonate SrCO 3 5.6 x 10 -10 Strontium fluoride SrF 2 4.3 x 10 -9 Strontium sulfate SrSO 4 3.4 x 10 -7 Zinc sulfide ZnS 2.0 x 10 -25

55. 14-6 Solubility Equilibria Write the chemical equation and expression for the solubility product constant for: a) Mg(OH) 2 b) Ca 3 (PO 4 ) 2 Answers: a) Mg(OH) 2 (s) ↔ Mg 2+ (aq) + 2OH - (aq) Ksp = [Mg 2+ ][OH - ] 2 b) Ca 3 (PO 4 ) 2 (s) ↔ 3Ca 2+ (aq) + 2PO 4 3- (aq) Ksp = [Ca 2+ ] 3 [PO 4 3- ] 2

56. 14-6 Solubility Equilibria When lead iodate, Pb(IO 3 ) 2, is added to water, a small amount dissolves. If measurements at 25°C show that the Pb 2+ concentration is 4.5 x 10 -5 M, calculate the value of K sp, for Pb(IO 3 ) 2 ? A: K sp = [Pb][IO 3 ] 2 = [4.5 x 10 -5 ][9.0 x 10 -5 ] 2 K sp = 3.6 x 10 -13

57. 14-6 Solubility Equilibria Calculate the solubility product for silver sulfate, given that the solubility is 0.44g/100mL. Ag 2 SO 4 MW = 311.8 g/mol [ ] = 1.4 x 10 -2 M Ag 2 SO 4 K sp = [Ag + ] 2 [SO 4 -2 ] = [2.8 x 10 -2 ] 2 [1.4 x 10 -2 ] = 1.1 x 10 -5

58. 14-6 Solubility Equilibria The solubility of a solid can be calculated from the solubility constant and the chemical equation for the dissociation of the solid.

59. Solving Solubility Problems For the salt AgI at 25  C, K sp = 1.5 x 10 -16 AgI(s)  Ag + (aq) + I - (aq) I C E O O +x x x 1.5 x 10 -16 = x 2 x = solubility of AgI in mol/L = 1.2 x 10 -8 M

60. Solving Solubility Problems For the salt PbCl 2 at 25  C, K sp = 1.6 x 10 -5 PbCl 2 (s)  Pb 2+ (aq) + 2Cl - (aq) I C E O O +x +2x x 2x 1.6 x 10 -5 = (x)(2x) 2 = 4x 3 x = solubility of PbCl 2 in mol/L = 1.6 x 10 -2 M

61. 14-6 Solubility Equilibria Given a K sp = 6.2 x 10 -12, calculate the solubility of La(IO 3 ) 3. A: La(IO 3 ) 3 (s)  La +3 (aq) + 3IO 3 -1 (aq) K sp = (x)(3x) 3 = 6.2 x 10 -12 x = 6.9 x10 -4 M

62. 14-7 Solubility & the Common Ion Effect Objectives Predict the solubility of a solid in a solution that contains a common ion Use approximations to calculate the roots of polynomial equations Decide whether a precipitate will form under a particular set of conditions

63. 14-7 Solubility & the Common Ion Effect How do we calculate the solubility of a precipitate in a solution that already contains one or more of the ions that compose the precipitate? The “common ion” acts to decrease the solubility of the precipitate. Note the relationship to Le Chatelier’s Principle!

64. Solving Solubility with a Common Ion For the salt AgI at 25  C, K sp = 1.5 x 10 -16 What is its solubility in 0.05 M NaI? AgI(s)  Ag + (aq) + I - (aq) I C E 0.05 O +x 0.05+x x 1.5 x 10 -16 = (x)(0.05+x)  (x)(0.05) x = solubility of AgI in mol/L = 3.0 x 10 -15 M

65. 14-7 Solubility & the Common Ion Effect What is the solubility of calcium hydroxide (K sp = 1.3 x 10 -6 ) in 0.050M sodium hydroxide solution? A: Ca(OH) 2 (s)  Ca +2 (aq) + 2OH -1 (aq) K sp = (x)(.05 + 2x) 2 = 1.3 x10 -6 M (Hint: Make an approximation that x is very small relative to 0.05) x = 5.2 x 10 -4 M

66. 14-7 Solubility & the Common Ion Effect Can we predict whether a precipitate will form when two solutions are mixed? Yes, we need to compare Q to K sp to determine the direction of the reaction. If Q is less than K sp then more solid can dissolve.

67. 14-7 Solubility & the Common Ion Effect A chemist mixes 200 mL of 0.010 M Pb(NO 3 ) 2 with 100 mL of 0.0050 M NaCl. Will lead chloride precipitate? (K sp = 1.6 x 10 -5 ) Remember lead chloride = PbCl 2 So Q = [Pb 2+ ][Cl 1- ] 2 = (0.0067)(0.0033) 2 = 7.4 x 10 -8 No, Q is less than the K sp.