1. by Steven S. Zumdahl & Don J. DeCoste University of Illinois Introductory Chemistry: A Foundation, 6 th Ed. Introductory Chemistry, 6 th Ed. Basic Chemistry, 6 th Ed.

2. Chapter 17 Equilibrium

3. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 3 Collision Theory of Kinetics Kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds. In order for a reaction to take place, the reacting molecules must collide into each other.

4. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 4 Collision Theory of Kinetics (cont.) Once molecules collide they may react together or they may not, depending on two factors: 1. Whether the reacting molecules collide in the proper orientation for new bonds to form 2. Whether the collision has enough energy to "break the bonds holding reactant molecules together"

5. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 5 Effective Collisions Collisions in which these two conditions are met (and the reaction occurs) are called effective collisions. The higher the frequency of effective collisions the faster the reaction rate.

6. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 6 Effective Collisions (cont.) When two molecules have an effective collision, a temporary, high energy (unstable) chemical species called an activated complex is formed. –Not a true molecule because its bonds aren’t complete

7. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 7 Activated Complex The difference in potential energy between the reactant molecules and the activated complex is called the activation energy, E a The larger the activation energy, the slower the reaction

8. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 8 Activated Complex (cont.) The energy to overcome the activation energy comes from the kinetic energy of the collision being converted into potential energy, or from energy available in the environment, i.e. heat. Different reactions have different activated complexes and therefore different activation energies.

9. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 9 Reactions and Activated Complex

10. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 10 Reactions and Activated Complex (cont.)

11. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 11 Factors Affecting Reaction Rate The kind of molecules and what condition the reactants are in Increasing temperature always increases reaction rate

12. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 12 Factors Affecting Reaction Rate (cont.)

13. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 13 The larger the concentration of reactant molecules, the faster the reaction will go. –Increases the frequency of reactant molecule collisions Factors Affecting Reaction Rate (cont.)

14. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 14 Factors Affecting Reaction Rate (cont.)

15. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 15 Catalysts are substances that affect the speed of a reaction without being consumed. Most catalysts are used to speed up a reaction. Homogeneous = catalyst present in same phase Heterogeneous = catalyst present in different phase Factors Affecting Reaction Rate (cont.)

16. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 16 Factors Affecting Reaction Rate (cont.)

17. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 17 Factors Affecting Reaction Rate (cont.)

18. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 18 Catalysts Speed Up Reactions Catalysts work by providing a pathway for the reaction with a lower activation energy. Enzymes are biochemical catalysts.

19. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 19 Catalysts Speed Up Reactions (cont.)

20. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 20 Increasing the temperature raises the kinetic energy of the reactant molecules, causing them to collide more frequently and to have more collisions with sufficient energy to form the activated complex. Molecular Interpretation of Factors Affecting Rate

21. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 21 The larger the concentration of reactant molecules, the more frequently they collide and the larger the number of effective collisions will be. Therefore the speed will increase. Molecular Interpretation of Factors Affecting Rate (cont.)

22. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 22 Catalysts work by providing an alternative pathway for the reaction with a lower activation energy. Lowering the activation energy means more molecules have enough kinetic energy so that when they collide they can form the activated complex. The result is the reaction goes faster. Molecular Interpretation of Factors Affecting Rate (cont.)

23. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 23 Reaction Dynamics If the products of a reaction are removed from the system as they are made, then a chemical reaction will proceed until the limiting reactants are used up. However, if the products are allowed to accumulate they will start reacting together to form the original reactants - called the reverse reaction.

24. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 24 Reaction Dynamics (cont.) The forward reaction slows down as the amounts of reactants decreases because the reactant concentrations are decreasing, while the reverse reaction speeds up as the concentration of the products increases. Eventually rate forward = rate reverse

25. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 25 Reaction Dynamics (cont.)

26. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 26 Chemical Equilibrium Equilibrium only occurs in a closed system. When a system reaches equilibrium, the amounts of reactants and products in the system stay constant.

27. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 27 Chemical Equilibrium (cont.)

28. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 28 Equilibrium Constant Law Of Chemical Equilibrium In this expression, K is a number called the equilibrium constant. aA + bB  cC + dD = [C] c [D] d [A] a [B] b K eq

29. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 29 Equilibrium Constant (cont.) Do not include solids or liquids, only solutions and gases. For a reaction, the value of K for a reaction depends on the temperature. K is independent of the amounts of reactants and products you start with. aA + bB  cC + dD = [C] c [D] d [A] a [B] b K eq

30. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 30 The relative concentrations of reactants and products when a reaction reaches equilibrium is called the position of equilibrium. Different initial amounts of reactants (and or products) will result in different equilibrium concentrations but the same equilibrium constant. Position of Equilibrium

31. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 31 If K is large, then there will be a larger concentration of products at equilibrium than of reactants, and we say the position of equilibrium favors the products. Position of Equilibrium (cont.)

32. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 32 If K is small then there will be a larger concentration of reactants at equilibrium than of products, and we say the position of equilibrium favors the reactants. The position of equilibrium is not affected by adding a catalyst. Position of Equilibrium (cont.)

33. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 33 Determine the value of the equilibrium constant for the reaction 2 SO 2 + O 2  2 SO 3 Example #1:

34. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 34 Determine the equilibrium expression Plug the equilibrium concentrations into the equilibrium expression Solve the equation 3.503.00SO 3 1.251.50O2O2 2.00SO 2 [Equilibrium][Initial]Chemical Example #1 (cont.)

35. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 35 Le Ch âtelier’s Principle Le Châtelier's Principle guides us in predicting the effect various changes have on the position of equilibrium. When a change is imposed on a system at equilibrium, the position of equilibrium will shift in the direction that will reduce the effect of that change.

36. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 36 The position of equilibrium can be affected without changing the equilibrium constant. Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found. Concentration Changes and Le Châtelier’s Principle

37. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 37 Concentration Changes and Le Châtelier’s Principle (cont.)

38. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 38 Changing the volume of a gas is like changing its concentration. –Has the same effect as changing the concentration on the position of equilibrium Decreasing the volume of the system increases its pressure. –Increasing the pressure on the system causes the position of equilibrium to shift toward the side of the reaction with the fewer gas molecules Changing Volume and Le Châtelier’s Principle

39. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 39 Changing Volume and Le Châtelier’s Principle (cont.)

40. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 40 The equilibrium constant will change if the temperature changes. For exothermic reactions, heating the system decreases K. –Think of heat as a product of the reaction –Therefore shift the position of equilibrium toward the reactants Changing Temperature and Le Châtelier’s Principle

41. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 41 Changing Temperature and Le Châtelier’s Principle (cont.) For endothermic reactions, heating the system increases K. –Think of heat as a reactant –The position of equilibrium will shift toward the products Cooling an exothermic or endothermic reaction will have the opposite effects on K and equilibrium position.

42. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 42 Example #2: If the value of the equilibrium constant for the reaction 2 SO 2 + O 2  2 SO 3 is 4.36, determine the equilibrium concentration of SO 3

43. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 43 Example #2 (cont.) Determine the equilibrium expression Plug the equilibrium concentrations and equilibrium constant into the equilibrium expression Solve the equation ?3.00SO 3 1.251.50O2O2 2.00SO 2 [Equilibrium][Initial]Chemical

44. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 44 Example #2 (cont.)

45. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 45 Solubility & Solubility Product Even “insoluble” salts dissolve somewhat in water –Insoluble = less than 0.1 g per 100 g H 2 O The solubility of insoluble salts is described in terms of equilibrium between undissolved solid and aqueous ions produced. A n X m (s)  n A + (aq) + m Y - (aq)

46. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 46 Solubility and Solubility Product (cont.) Equilibrium constant called solubility product. K sp = [A + ] n [Y - ] m If undissolved solid is in equilibrium with the solution, the solution is saturated. Larger K = more soluble –For salts that produce same the number of ions

47. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 47 Example #3: Calculate the solubility of AgI in water at 25°C if the value of K sp = 1.5 x 10 -16.

48. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 48 Example #3 (cont.) Determine the balanced equation for the dissociation of the salt AgI(s)  Ag + (aq) + I - (aq) Determine the expression for the solubility product –Same as the equilibrium constant expression K sp = [Ag + ][I - ]

49. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 49 Define the concentrations of dissolved ions in terms of x AgI(s)  Ag + (aq) + I - (aq) Stoichiometry tells us that we get 1 mole of Ag + and 1 mol I - for each mole of AgI dissolved. Let x = [Ag + ], then [I - ] = x Plug the ion concentrations into the expression for the solubility product and solve for K sp [Ag + ] = [I - ] = x Example #3 (cont.)

50. Copyright © Houghton Mifflin Company. All rights reserved. 17 | 50 [Ag + ] = 1.2 x 10 -8 mol/L = [AgI] The solubility of AgI = 1.2 x 10 -8 M Example #3 (cont.)