1. Foundations of College Chemistry, 14 th Ed. Morris Hein and Susan Arena Keeping fish in an aquarium requires maintaining an equilibrium among the living organisms and the water. 16 Chemical Equilibrium Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

2. 16.1 Rates of Reaction 16.2 Chemical Equilibrium 16.3 Le Châtelier’s Principle 16.4 Equilibrium Constants 16.5 Ion Product Constant for Water 16.6 Ionization Constants 16.7 Solubility Product Constant 16.8 Buffer Solutions: The Control of pH Chapter Outline © 2014 John Wiley & Sons, Inc. All rights reserved.

3. Chemical Kinetics: study of reaction rates and reaction mechanisms. Factors That Affect Reaction Rates 1. Frequency of collisions between reactants (concentration effects). 2. Energy needed for effective collisions between reactants to produce products (temperature and catalytic effects). Rates of Reaction © 2014 John Wiley & Sons, Inc. All rights reserved.

4. Most chemical reactions are reversible, consisting of a forward reaction (where reactants are converted to products) and a reverse reaction (where products are converted back to reactants). A + B C + D Eventually, the rate of the forward reaction is equal to the rate of the reverse reaction. This point is when equilibrium is attained. C + D A + B (forward reaction) (reverse reaction) A + B C + D Reversible Reactions © 2014 John Wiley & Sons, Inc. All rights reserved.

5. Equilibrium vapor pressures are measured at different temperatures to generate vapor pressure curves. liquid + heat vapor (evaporation) liquid + heat vapor Forward Reaction: Reverse Reaction: vapor liquid + heat (condensation) At equilibrium: Rate of evaporation = Rate of condensation At this point, the vapor pressure of the liquid does not change with time. © 2014 John Wiley & Sons, Inc. All rights reserved. Reversible Reactions

6. 2 NO 2 (g) N 2 O 4 (g) + heat Brown gas Colorless gas N 2 O 4 (g) + heat 2 NO 2 (g) 25 °C90 °C Reversible reaction of NO 2 and N 2 O 4. More dark brown NO 2 molecules are present at higher temperature. Forward Reaction: Reverse Reaction: © 2014 John Wiley & Sons, Inc. All rights reserved. Reversible Reactions

7. The rate of the forward reaction (red) and back reaction (blue) become equal at equilibrium (purple). The forward reaction rate decreases as reactants are consumed to form products. The reverse reaction rate starts at 0 (no product is present to react at this time) and increases as the amount of product increases. Rates of Reactions and Equilibrium © 2014 John Wiley & Sons, Inc. All rights reserved.

8. Chemical Equilibrium: a dynamic state in which two opposing processes (forward and reverse reactions) occur simultaneously at the same rate. When chemical equilibrium is achieved: Rate forward reaction = Rate reverse reaction HF (aq) + H 2 O (l) H 3 O + (aq) + F – (aq) Example At equilibrium, HF is ionizing at the same rate that the acid is reforming, so the concentrations of HF, H 3 O + and F – remain constant. Chemical Equilibrium © 2014 John Wiley & Sons, Inc. All rights reserved.

9. Equilibrium is achieved in a chemical reaction when: a. Reactants are completely consumed. b. The concentrations of all reactants and products become equal. c. The rates of the opposing reactions become equal. d. The forward and reverse reactions stop. Chemical Equilibrium Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

10. Le Châtlelier’s principle: if a stress is applied to a system, the system will respond to relieve that stress and restore equilibrium under the new set of conditions. Common Stressors of Chemical Equilibria 1. Changes in concentration 2. Temperature changes 3. Changes in gas volume Le Châtelier’s Principle © 2014 John Wiley & Sons, Inc. All rights reserved.

11. For the reaction: 3 H 2 (g) + N 2 (g) 2 NH 3 (g) At equilibrium: Rate forward reaction = Rate reverse reaction If H 2 is added to the reaction at equilibrium, the forward rate would be increased, producing more NH 3 and consuming additional H 2 and N 2. As the amount of NH 3 increases, the rate of the forward reaction will decrease while the reverse reaction rate increases (see Slide 16-7). Eventually, the two rates become equal again and equilibrium is achieved. Effect of Concentration © 2014 John Wiley & Sons, Inc. All rights reserved.

12. If H 2 is added to the reaction at equilibrium, more NH 3 is produced. The equilibrium is said to have shifted to the right. Summary of H 2 Addition to the System CompoundChange in Reagent Concentration H2H2 decreases N2N2 NH 3 increases For the reaction: 3 H 2 (g) + N 2 (g) 2 NH 3 (g) Adding Reactant to a System at Equilibrium © 2014 John Wiley & Sons, Inc. All rights reserved.

13. For the generic reaction: Reactant Product The following table summarizes effects of added/removed reagents on the equilibrium. CompoundShiftIncrease in equilibrium concentration Decrease in equilibrium concentration Add reactantRightProductReactant Remove reactantLeftReactantProduct Add productLeftReactantProduct Remove productRightProductReactant © 2014 John Wiley & Sons, Inc. All rights reserved. Adding Reactant to a System at Equilibrium

14. For the reaction: Cu 2+ (aq) + 4 NH 3 (aq) [Cu(NH 3 ) 4 2+ ] (aq) What color will be observed if ammonia is added to an equilibrium mixture? Ammonia will shift the reaction to the right, resulting in a royal blue color. Pale blueRoyal blue Summary of NH 3 Addition to the System CompoundChange in Reagent Concentration Cu 2+ decreases NH 3 decreases NH 3 increases Effect of Concentration © 2014 John Wiley & Sons, Inc. All rights reserved.

15. For the reaction: CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) Adding CH 3 COO - to the equilibrium mixture will: 1. Decrease the concentration of CH 3 COO -. 2. Cause an increase in the rate of the reverse reaction, shifting the reaction to the left. 3. Decrease the H 3 O +, increasing the pH. Effect of added CH 3 COO - concentration on observed pH © 2014 John Wiley & Sons, Inc. All rights reserved. Effect of Concentration

16. Adding NaOH to the equilibrium mixture will: 1. Decrease the H 3 O +, as the hydronium ion would react with the hydroxide ion. 2. Would cause a decrease in the reverse reaction rate, causing the reaction to shift towards the right. CompoundChange in Concentration CH 3 COOHdecreases H3O+H3O+ CH 3 COO - increases Summary of NaOH Addition to the System For the reaction: CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) © 2014 John Wiley & Sons, Inc. All rights reserved. Effect of Concentration

17. For the reaction: In which direction will the equilibrium shift when the concentration of nitrogen is increased? a.Shift to the right b.Shift to the left c.No shift will occur 3 H 2 (g) + N 2 (g) 2 NH 3 (g) Summary of N 2 Addition to the System CompoundChange in Reagent Concentration H2H2 decreases N2N2 NH 3 increases Effect of Concentration Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

18. For the reaction: In which direction will the equilibrium shift when the concentration of chloride ion is increased? a.Shift to the right b.Shift to the left c.No shift will occur AgCl (s) Ag + (aq) + Cl - (aq) Summary of Cl - Addition to the System CompoundChange in Reagent Concentration AgClincreases Ag + decreases Cl - decreases © 2014 John Wiley & Sons, Inc. All rights reserved. Effect of Concentration Practice

19. What color will the reaction be if HCl is added, which will react with the ammonia? a.Pale blue b.Royal blue c.No color change Summary of HCl Addition to the System CompoundChange in Reagent Concentration NH 3 increases Cu 2+ increases Cu(NH 3 ) 4 2+ decreases For the reaction: Cu 2+ (aq) + 4 NH 3 (aq) [Cu(NH 3 ) 4 2+ ] (aq) Pale blueRoyal blue © 2014 John Wiley & Sons, Inc. All rights reserved. Effect of Concentration Practice

20. For the reaction: If I 2 (g) is added to the equilibrium mixture, the concentration of H 2 will: a.Increase b.Decrease c.Not change H 2 (g) + I 2 (g) 2 HI (g) Summary of I 2 Addition to the System CompoundChange in Reagent Concentration H2H2 decreases I2I2 increases HIincreases © 2014 John Wiley & Sons, Inc. All rights reserved. Effect of Concentration Practice

21. A decrease in volume in a gas phase reaction will increase the pressure of all gases (both reactants and products). The balanced equation determines whether the change in volume will cause a shift to the left or right. The reaction will shift to the side with the smaller number of gas molecules when the volume is decreased. The reaction will shift to the side with the larger number of gas molecules when the volume is increased. Effect of Changes in Volume © 2014 John Wiley & Sons, Inc. All rights reserved.

22. For the reaction: 3 H 2 (g) + N 2 (g) 2 NH 3 (g) Summary of Volume Change in the System CompoundChange in Reagent Concentration H2H2 decreases N2N2 NH 3 increases 4 moles of gas 2 moles of gas How will a decrease in the volume of the container affect the equilibrium concentrations? The equilibrium will shift to the right, producing more ammonia. © 2014 John Wiley & Sons, Inc. All rights reserved. Effect of Changes in Volume

23. For the reaction: H 2 (g) + I 2 (g) 2 HI (g) 2 moles of gas 2 moles of gas The equilibrium will remain unchanged, as the number of gas molecules is the same on both sides of the chemical equation. © 2014 John Wiley & Sons, Inc. All rights reserved. Effect of Changes in Volume

24. For the reaction: In which direction will the equilibrium shift when the volume of the reaction vessel is decreased? a.Shift to the right b.Shift to the left c.No change occurs Summary of Volume Decrease CompoundChange in Reagent Concentration PCl 5 increases PCl 3 decreases Cl 2 decreases PCl 5 (g) PCl 3 (g) + Cl 2 (g) 1 mole of gas 2 moles of gas Effect of Volume Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

25. For the reaction: In which direction will the equilibrium shift when the volume of the reaction vessel is increased? a.Shift to the right b.Shift to the left c.No change occurs Summary of Volume Increase CompoundChange in Reagent Concentration CO 2 decreases COincreases O2O2 2 CO 2 (g) 2 CO (g) + O 2 (g) 2 moles of gas 3 moles of gas © 2014 John Wiley & Sons, Inc. All rights reserved. Effect of Volume Practice

26. For the reaction: In which direction will the equilibrium shift when the volume of the reaction vessel is decreased? a.Shift to the right b.Shift to the left c.No change occurs AgCl (s) Ag + (aq) + Cl - (aq) 0 moles of gas 0 moles of gas Remember, pressure and volume changes only affect reactions involving gases! © 2014 John Wiley & Sons, Inc. All rights reserved. Effect of Volume Practice

27. An increase in temperature increases both the rates of the forward and reverse reactions because of the increase in kinetic energy of all collisions in the system. The application of heat to increase the temperature favors a reaction where heat is a reactant (i.e. heat is absorbed). When heat is absorbed, the reaction is called endothermic. A + heat B A B + heat When heat is given off, the reaction is called exothermic. Treat heat as any other reactant or product in a Le Châtelier problem to predict effects on equilibrium! Effect of Temperature © 2014 John Wiley & Sons, Inc. All rights reserved.

28. Increasing the reaction temperature favors the reverse reaction, shifting the equilibrium to the left towards the brown gas. For the reaction: 2 NO 2 (g) N 2 O 4 (g) + heat Brown gasColorless gas 25°C90°C Reversible reaction of NO 2 and N 2 O 4. More dark brown NO 2 molecules are present at higher temperature. © 2014 John Wiley & Sons, Inc. All rights reserved. Effect of Temperature

29. How will an increase in temperature affect the equilibrium concentration of ammonia? For the reaction: 3 H 2 (g) + N 2 (g) 2 NH 3 (g) + heat 1. The reaction is exothermic (heat is evolved). 2. Increasing the temperature adds heat to the system. 3. The reverse reaction is favored and the equilibrium will shift to the left. 4. The amount of ammonia will then decrease. © 2014 John Wiley & Sons, Inc. All rights reserved. Effect of Temperature

30. For the reaction: In which direction will the equilibrium shift when the reaction is cooled? a.Shift to the right b.Shift to the left c.No change occurs Summary of Temperature Decrease CompoundChange in Reagent Concentration PCl 5 increases PCl 3 decreases Cl 2 decreases PCl 5 (g) + heat PCl 3 (g) + Cl 2 (g) Effect of Temperature Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

31. Catalyst: substance that increases the rate of a chemical reaction but that can be fully recovered at the end of the reaction. A catalyst does not shift the equilibrium of a chemical reaction. A catalyst lowers the activation energy of a reaction, affecting only the rate of reaction. Activation energy: minimum energy required for a chemical reaction to occur. Effect of Catalysts © 2014 John Wiley & Sons, Inc. All rights reserved.

32. Energy Diagram for an Exothermic Reaction Reaction Coordinate Diagram © 2014 John Wiley & Sons, Inc. All rights reserved.

33. For the reaction: In which direction will the equilibrium shift when a catalyst is added? a.Shift to the right b.Shift to the left c.No change occurs 3 H 2 (g) + N 2 (g) 2 NH 3 (g) + heat Catalysts only affect the rate of reaction, not the chemical equilibrium! Effect of Catalyst Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

34. For the equilibrium: a A + b B c C + d D Equilibrium constant = K eq = [C] c [D] d [A] a [B] b There is a mass law expression defined as the equilibrium constant (K eq ): Only substances which possess molar concentrations that vary will appear in the equilibrium constant expression. Gases and aqueous solutions are the only substances that typically appear in equilibrium constant expressions. Equilibrium Constants © 2014 John Wiley & Sons, Inc. All rights reserved.

35. For the equilibrium: The equilibrium constant can be written as: The value of K eq is determined by the concentrations of both the reactants and products. 3 H 2 (g) + N 2 (g) 2 NH 3 (g) K eq = [NH 3 ] 2 [H 2 ] 3 [N 2 ] © 2014 John Wiley & Sons, Inc. All rights reserved. Equilibrium Constants

36. The magnitude of an equilibrium constant indicates the extent of a chemical reaction. K eq >> 1 indicates the relative amounts of products are favored when compared to the reactants. K eq << 1 indicates the relative amounts of reactants are favored when compared to the products. K eq ~ 1 means that both reactants and products are present in significant amounts. © 2014 John Wiley & Sons, Inc. All rights reserved. Equilibrium Constants

37. Calculate K eq for the following reaction: when [PCl 5 ] = 0.030 M, [PCl 3 ] = 0.97 M and [Cl 2 ] = 0.97 M. (0.97)(0.97) 0.030 K eq = [PCl 5 ] = = 31 PCl 5 (g) PCl 3 (g) + Cl 2 (g) [PCl 3 ][Cl 2 ] © 2014 John Wiley & Sons, Inc. All rights reserved. Equilibrium Constants

38. Calculate K eq for the following reaction: H 2 (g) + I 2 (g) 2 HI (g) when [H 2 ] = 0.228 M, [I 2 ] = 0.228 M and [HI] = 1.544 M. a.29.7 b.59.4 c.0.0337 d.0.0219 e.45.9 (1.544) 2 (0.228)(0.228) K eq = [HI] 2 [H 2 ][I 2 ] = = 45.9 Equilibrium Constants Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

39. Pure water auto(self) ionizes [H 3 O + ] = [OH - ] = 1.00 x 10 -7 M H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) Ion Product Constant for Water Realize any water solution contains both H 3 O + and OH -. K w = [H 3 O + ][OH - ] = 1.00 x 10 -14 Ion Product Constant for Water © 2014 John Wiley & Sons, Inc. All rights reserved.

40. K w = [H 3 O + ][OH - ] = 1.00 x 10 -14 Ion Product Constant for Water pH = -log[H 3 O + ] pOH = -log[OH - ] pH and pOH pH + pOH = 14 Relationship Between pH and pOH Relationship Between [H 3 O + ] and [OH - ] and pH/pOH Relationship Between [H 3 O + ] and [OH – ] © 2014 John Wiley & Sons, Inc. All rights reserved.

41. What is the [H 3 O + ] in a 0.0152 M solution of NaOH? pH =- log[H 3 O + ] [H 3 O + ] = 1.00 x 10 -14 /[OH - ] K w = [H 3 O + ][OH - ] = 1.00 x 10 -14 Calculate the pH of a 0.0152 NaOH solution. = 1 x 10 -14 /0.0152 = 6.58 x 10 -13 M = -log(6.58 x 10 -13 ) = 12.182 Using K w and pH Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

42. What is the [OH - ] in a 0.00010 M solution of HCl? [OH - ] = 1.00 x 10 -14 /[H 3 O + ] K w = [H 3 O + ][OH - ] = 1.00 x 10 -14 a.1 x 10 -14 M b.1 x 10 -10 M c.1 x 10 -4 M d.1 x 10 -7 M What is the pOH of a 0.00010 M solution of HCl? a.1 b.4 c.10 d.13 pOH = -log[OH - ] = 1 x 10 -14 /0.0001 = 1.00 x 10 -10 M = -log(1.00 x 10 -10 ) = 10.0 © 2014 John Wiley & Sons, Inc. All rights reserved. Using K w and pH Practice

43. Acid Ionization Constant (K a ): a measure of the extent to which a weak acid ionizes in water. Water is the solvent. Because its concentration does not change measurably during the ionization, water is not included in the K a expression. HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) For a general weak acid (HA): K a = [H 3 O + ][A - ] [HA] The larger the value of K a, the more significant the ionization, the stronger the acid. Ionization Constants © 2014 John Wiley & Sons, Inc. All rights reserved.

44. Determine the [H 3 O + ] of a 0.20 M solution of CH 3 COOH. Equation CH 3 COOH + H 2 O CH 3 COO - + H 3 O + Initial (M) 0.2 0 0 Change (M) - x + x + x Equilibrium (M) 0.2 - x x x K a = [CH 3 COO - ][H 3 O + ] [CH 3 COOH] = 1.8 x 10 -5 = x2x2 0.20 - x x = [H 3 O + ] = 1.90 x 10 -3 M K a = 1.8 x 10 -5 x 2 = (0.20 - x)(1.8 x 10 -5 )x 2 = 3.6 x 10 -6 - 1.8 x 10 -5 x Calculating [H 3 O + ] in a Weak Acid © 2014 John Wiley & Sons, Inc. All rights reserved.

45. Determine the % ionization of a 0.20 M CH 3 COOH solution. 1.9 x 10 -3 0.20 CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) % ionization = [H 3 O + ] initial acid concentration x 100 % % ionization = x 100% = 95% % Ionization of a Weak Acid © 2014 John Wiley & Sons, Inc. All rights reserved.

46. Determine the [H 3 O + ] of a 0.10 M solution of HCN. Equation HCN + H 2 O CN - + H 3 O + Initial (M) 0.10 0 0 Change (M) - x + x + x Equilibrium (M) 0.10 - x x x K a = [CN - ][H 3 O + ] [HCN] = 4.0 x 10 -10 = x2x2 0.10 - x x = [H 3 O + ] = 6.3 x 10 -6 M K a = 4.0 x 10 -10 x 2 = (0.10 - x)(4.0 x 10 -10 ) x 2 = 4.0 x 10 -11 - 4.0 x 10 -10 x Calculating [H 3 O + ] in a Weak Acid © 2014 John Wiley & Sons, Inc. All rights reserved.

47. Saturated solutions have solid in equilibrium with dissolved solute. The solubility product (K sp ) is defined as: The amount of solid does not affect the equilibrium and is not included in the equilibrium expression. A x B y (s) x A y+ (aq) + y B x- (aq) K sp = [A y+ ] x [B x- ] y Solubility Product Constants © 2014 John Wiley & Sons, Inc. All rights reserved.

48. Calculate the solubility (x), [Hg 2+ ], and [Br - ] of HgBr 2 if K sp = 1.3 x 10 -19. K sp = [Hg 2+ ][Br - ] 2 = 1.3 x 10 -19 Equation HgBr 2 (s) Hg 2+ (aq) + 2 Br - (aq) Initial (M) 0 0 Change (M) +x +2x Equil. (M) x 2x K sp = x(2x) 2 = 4x 3 x = solubility = [Hg 2+ ] = 3.2 x 10 -7 M [Br - ] = 2x = 6.4 x 10 -7 M Solubility Product Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

49. The K sp of AgI is 8.3 x 10 -17. What is the solubility of AgI? K sp = [Ag + ][I - ] = 8.3 x 10 -17 Equation AgI (s) Ag + (aq) + I - (aq) Initial (M) 0 0 Change (M) +x +x Equil. (M) x x K sp = x(x) = x 2 x = solubility = [Ag + ] = [I - ] = (8.3 x 10 -17 ) 0.5 = 9.1 x 10 -9 a.8.3 x 10 -17 b.1.7 x 10 -16 c.2.7 x 10 -6 d.9.1 x 10 -9 © 2014 John Wiley & Sons, Inc. All rights reserved. Solubility Product Practice

50. Common ion effect: a shift in the equilibrium, based on Le Châtelier’s principle, when additional ion already present in a solution is added. Example Sodium hydroxide is added to a saturated solution of magnesium hydroxide until the [OH - ] is 0.010 M. Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH - (aq) The addition of hydroxide ions shifts the equilibrium to the left, reducing the magnesium ions in solution. Common Ion Effect © 2014 John Wiley & Sons, Inc. All rights reserved.

51. Equation Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH - (aq) Initial (M) 0 0 Change (M) +x 0.010 Equil. (M) x 0.010 K sp = x(0.010) = 5.6 x 10 -12 What is the [Mg 2+ ] in solution when sodium hydroxide is added to a saturated Mg(OH) 2 solution until the [OH - ] = 0.010 M? (K sp = 5.6 x 10 -12 ) x = 5.6 x 10 -12 /0.010 = 5.5 x 10 -8 M Common Ion Effect Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

52. Buffer solution: resists changes in pH when diluted or small amounts of acid or base are added. Buffers are prepared by mixing together (usually equimolar amounts) of: 1. A weak acid with a salt containing its conjugate base. 2. A weak base with a salt containing its conjugate acid. Buffer capacity: extent to which a buffer can absorb added acid or base and still maintain the pH. Buffer Solution © 2014 John Wiley & Sons, Inc. All rights reserved.

53. Consider a buffer prepared using 0.1 M CH 3 COOH and 0.1 M NaOOCCH 3. The solution contains an acid (CH 3 COOH) which can neutralize base when added, so the pH does not change. The solution also contains a base (CH 3 COO - ) which can neutralize acid when added, so the pH does not change. CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) CH 3 COOH (aq) + H 2 O (l) Buffers © 2014 John Wiley & Sons, Inc. All rights reserved.

54. Buffers

55. List the factors that affect the rate of a chemical reaction. 16.1 Rates of Reaction Define a reversible chemical reaction and explain what is occurring in a chemical reaction at equilibrium. 16.2 Chemical Equilibrium Use Le Châtelier’s principle to predict the changes that occur when concentration, temperature or volume is changed in a system at equilibrium. 16.3 Le Châtelier’s Principle Learning Objectives © 2014 John Wiley & Sons, Inc. All rights reserved.

56. Write the general expression for the equilibrium constant for a reaction. 16.4 Equilibrium Constants Calculate the concentrations of H + and OH - in a solution using the ion product constant for water. 16.5 Ion Product Constant for Water Use the ionization constant of a reactant in an equilibrium expression to find the percent ionization of a substance in solution and to find the pH of a weak acid. 16.6 Ionization Constants © 2014 John Wiley & Sons, Inc. All rights reserved. Learning Objectives

57. Use the solubility product constant to calculate the solubility of a slightly soluble salt and to determine whether a precipitate will form in a solution. 16.7 Solubility Product Constant © 2014 John Wiley & Sons, Inc. All rights reserved. Learning Objectives