1. Chemical Equilibrium Chapter 13

2. Equilibrium: the extent of a reaction In stoichiometry we talk about theoretical yields, and the many reasons actual yields may be lower. A critical reason actual yields may be lower is the reversibility of chemical reactions. Equilibrium looks at the extent of a chemical reaction.

3. Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when:  the rates of the forward and reverse reactions are equal and  the concentrations of the reactants and products remain constant Physical equilibrium (between states of matter) H 2 O (l) Chemical equilibrium N2O4 (g)N2O4 (g) 14.1 H 2 O (g) 2NO 2 (g)

4. The Concept of Equilibrium Consider colorless frozen N 2 O 4; at room temperature, it decomposes to brown NO 2 : N 2 O 4 (g)  2NO 2 (g). At some time, the color stops changing and we have a mixture of N 2 O 4 and NO 2. (chemical equilibrium) Using the collision model: –as the amount of NO 2 builds up, there is a chance that two NO 2 molecules will collide to form N 2 O 4. –At the beginning of the reaction, there is no NO 2 so the reverse reaction (2NO 2 (g)  N 2 O 4 (g)) does not occur.

5. 5 COLLISION MODEL Consider colorless frozen N 2 O 4; at room temperature, it decomposes to brown NO 2 : N 2 O 4 (g)  NO 2 (g) Dimerization Forward reaction As the substance warms it begins to decompose: N 2 O 4 (g)  2NO 2 (g) Reverse reaction When enough NO 2 is formed, it can react to form N 2 O 4 : 2NO 2 (g)  N 2 O 4 (g).

6. As long as there is N 2 O 4 present, the forward reaction will keep on happening and as long as there is NO 2 present, the reverse reaction will keep on happening! In any reversible reaction, the forward reaction and the reverse reactionare going on at the same time! On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

7. N 2 O 4 (g) 2NO 2 (g) Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4 equilibrium 14.1

8. Characteristics of a System at Dynamic Equilibrium 1.The rate of the forward reaction = The rate of the reverse reaction 2.Microscopic processes (the forward and reverse reaction) continue in a balance which yields no macroscopic changes. (so nothing appears to be happening.) 3.The system is closed and the temperature is constant and uniform throughout. 4.The equilibrium can be approached from the left (starting with reactants) or from the right (starting with products).

9. N 2 O 4 (g) 2NO 2 (g) = 4.63 x 10 -3 K = [NO 2 ] 2 [N 2 O 4 ] jA +kB lC + mD K = [C] l [D] m [A] j [B] k Law of Mass Action: Description of the equilibrium condition. K >> 1 K << 1 Lie to the rightFavor products Lie to the left Favor reactants Equilibrium Will 14.1 The Equilibrium Constant The Equilibrium Constant No matter the starting composition of reactants and products,the same ratio of concentrations is achieved at equilibrium.

10. 10 Mass action applies to solutions and gases. NH 3 synthesis at 500°C Same regardless the amount of gasses mixed. Although the ratio of product to reactant is constant for a given system at a given T, the [equilibrium] will not always be the same.

11. 11 Each set of [equilibrium] is called: Equilibrium position Can be an infinite # of equilibrium positions Depends on initial concentrations

12. 12 Equilibrium Expression Write the equilibrium expression for: 4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O(g)

13. The Equilibrium Expression Write the equilibrium expression for the following reaction: k = [NH 3 ] 2 [N 2 ][H 2 ] 3

14. The following equilibrium concentrations were observed for the Haber process at 127°C: [NH 3 ] = 3.1 x 10 -2 mol/L [N 2 ] = 8.5 x 10 -1 mol/L [H 2 ] = 3.1 x 10 -3 mol/L a.Calculate the value of K at 127°C for this reaction. b.Calculate the value of the equilibrium constant at 127°C for the reaction. 2NH 3  N 2 + 3H 3 c.And this reaction: ½ N 2 + 3/2 H 2  NH 3

15. 15 If the reaction is reversed the new equilibrium expression is the reciprocal of. K = [NH 3 ] 2 [N 2 ][H 2 ] 3 K = (3.1 x 10 -2 mol/L) 2 (8.5 x 10 -1 mol/L)(3.1 x 10 -2 mol/L) 3 = 3.8 x 10 4 (no units) a. b.K = [NH 3 ] 2 [N 2 ][H 2 ] 3 K = 1 K = 1/3.8 x 10 4 = 2.6 x 10 -5

16. 16 K = [NH 3 ] [N 2 ] 1/2 [H 2 ] 3/2 K = (3.1 x 10 -2 mol/L) 2 (8.5 x 10 -1 mol/L)(3.1 x 10 -2 mol/L) 3 = 3.8 x 10 4 c. [NH 3 ] 2 [N 2 ][H 2 ] 3 K = 1/2 K If the original reaction is multiplied by some factor n, the equilibrium expression is raised to the n th power. ½ N 2 + 3/2 H 2  NH 3 = () 1/2 = (3.8 x 10 4 ) 1/2 = 1.9.x 10 2 ½ (N 2 + 3H 2  2NH 3 )

17. A + B C + D C + D E + F A + B E + F K c = ‘ [C][D] [A][B] K c = ‘ ‘ [E][F] [C][D] [E][F] [A][B] K c = KcKc ‘ KcKc ‘‘ KcKc KcKc ‘‘ KcKc ‘ x If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2 MULTIPLE EQUILIBRIA RULE

18. 18 Equilibrium Expressions Involving Pressures PV=nRT

19. 19 Also write in terms of partial pressure K and Kc =[equilibrium] Used interchangeably

20. 20 The relationship between K and K p for ideal gases, C = P/RT

21. Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N 2 O 4 (g) 2NO 2 (g) K = [NO 2 ] 2 [N 2 O 4 ] K p = NO 2 P2P2 N2O4N2O4 P In most cases K  K p aA (g) + bB (g) cC (g) + dD (g) 14.2 K p = K(RT)  n  n = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

22. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = 0.012 M, [Cl 2 ] = 0.054 M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K and K p. CO (g) + Cl 2 (g) COCl 2 (g) K = [COCl 2 ] [CO][Cl 2 ] = 0.14 0.012 x 0.054 = 220 K p = K(RT)  n  n = 1 – 2 = -1 R = 0.0821T = 273 + 74 = 347 K K p = 220 x (0.0821 x 347) -1 = 7.7 14.2

23. 23 Heterogeneous Equilibria... are equilibria that involve more than one phase. CaCO 3 (s)  CaO(s) + CO 2 (g) K = [CO 2 ] The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.

24. 24 The position of the equilibrium does not depend on the amounts of CaCo 3 (s) and CaO(s) ***Concentration of pure solids and liquids can not change.

25. Extent of a reaction K> 1 products are favored K< 1 reactants are favored the size of K and the time required to reach equilibrium are not directly related. Time: reaction rate; size of activation energy Size of K: thermodynamic (∆E of reactants and products.

26. 26 Reaction Quotient... helps to determine the direction of the move toward equilibrium. Apply the law of mass action to the initial concentrations instead of equilibrium

27. 27 Reaction Quotient (continued) H 2 (g) + F 2 (g)  2HF(g) Subscript indicates initial concentrations.

28. The reaction quotient (Q c ) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (K c ) expression. IF Q c < K c system shifts to the right Q c = K c the system is at equilibrium Q c > K c system shifts to the left 14.4

29. 29 For the synthesis of ammonia at 500C, the equilibrium constant is 6.0 x 10 -2. Predict the direction of the shift if: [NH 3 ] 0= 1.0x10 -3 M [N 2 ] 0 =1.0x10 -5 M [H 2 ] 0 =2.0x10 -3 M

30. 30 Calculating Equilibrium Pressures and Concentrations Dinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar lander of the NASA Apollo missions. In the gas phase it decomposes to gaseous nitrogen dioxide: N 2 O 4 (g) ↔2NO 2 (g) Consider an experiment in which gaseous N 2 O 4 was placed in a flask and allowed to reach equilibrium at a temperature where Kp= 0.133. At equilibrium, the pressure of N 2 O 4 was found to be 2.71atm. Calculate the equilibrium pressure of NO 2 (g)

31. 31 We know that the Kp of the gases NO 2 and N 2 O 4 must satisfy the relationship Kp= Since we know P NO2 2 = Kp(P N2O4 ) = (0.133)(2.71)= 0.360 Therefore, P NO2 = √0.360 = 0.600

32. 32 At a certain temperature a 1.00L flask initially contained 0.298 mol PCl 3 (g) and 8.7x10 -3 mol PCl 5 (g). After the system had reached equilibrium, 2.00x10 -3 mol Cl 2 (g) was found in the flask. Gaseous PCl 5 decomposes according to the reaction PCl 5 (g)↔PCl 3 (g) + Cl 2 (g) Calculate the equilibrium concentrations of all species and the value of K.

33. 33 1.What is the equilibrium expression? 2. Begin with initial concentrations. [Cl 2 ] = 0 [PCl 3 ] =.298 M [PCl 5 ] = 8.7 x 10 -3 M 3. Find the change required to reach equilibrium. PCl 5 (g) ↔ PCl 3 (g) + Cl 2 (g) 2.0 x 10 -3 mol  2.0 x 10 -3 mol + 2.0 x 10 -3 mol ** reaction shifts to the right**

34. 4. Apply this change to the initial concentrations. [Cl 2 ] = 0 +2.0 x 10 -3 M [PCl 3 ] =.298 M + 2.0 x 10 -3 M =.300 M [PCl 5 ] = 8.7 x 10 -3 M - 2.0 x 10 -3 M = 6.7 x 10 -3 M 5. Use [equilibrium] to find K. 8.96 x 10 -2

35. Writing Equilibrium Constant Expressions The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. The equilibrium constant: no units In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

36. 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 You will use it for the next 3 chapters!

37. 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:

38. Solving for Equilibrium Concentration H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) Step #2: We “ICE” the problem, beginning with the Initial concentrations 8600 -x +x 8-x6-x x x Step #3: Find Q, determine the direction of the shift.(change)

39. Solving for Equilibrium Concentration Step #4: We plug equilibrium concentrations into our equilibrium expression, and solve for x H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) x = 4

40. 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) x = 4

41. The reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.1.5 x 10 2 at a certain temperature. If 3.00 mol of each component was added to a 1.50 L flask, calculate the equilibrium concentrations of all species. H 2 (g) + F 2 2HF (g) Let x be the change in concentration of H 2 and F 2 Initial (M) Change (M) Equilibrium (M) 2.00 -x-x +2x 2.00 - x2.00 + 2x [HF] 2 [H 2 ][F 2 ] K = (2.00 + 2x) 2 = 1.15 x 10 2 Solve for x 14.4 2.00 [HF] 2 [H 2 ][F 2 ] Q = Shift to the right -x-x 2.00 - x (2.00 - x) 2 X = 1.528

42. At 1280 0 C the equilibrium constant (K) for the reaction Is 1.1 x 10 -3. If the initial concentrations are [Br 2 ] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x [Br] 2 [Br 2 ] K = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 Solve for x 14.4

43. K = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 4x 2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x 2 + 0.0491x + 0.0000747 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x x = -0.00178x = -0.0105 At equilibrium, [Br] = 0.012 + 2x = -0.009 Mor 0.00844 M At equilibrium, [Br 2 ] = 0.062 – x = 0.0648 M 14.4

44. Approximating If Keq is really small the reaction will not proceed to the right very far, meaning the equilibrium concentrations will be nearly the same as the initial concentrations of your reactants. 0.20 – x is just about 0.20. since x is really small. If the difference between Keq and initial concentrations is around 3 orders of magnitude or more, go for it. Otherwise, you have to use the quadratic.

45. Example Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: 0.20 0 -x +2x 0.20-x 2x With an equilibrium constant that small, whatever x is, it’s near nill, and 0.20 minus nill is 0.20 (like a million dollars minus a nickel is still a million dollars). 0.20 – x is the same as 0.20 x = 3.83 x 10 -6 M More than 3 orders of mag. between these numbers. The simplification will work here.

46. How valid is the approximation? x = 3.83 x 10 -6 M 0.20 – x =.199996 3.83 x 10 -6 ÷.20 =.000015 =.0015% Determine the % difference to find the validity.

47. 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 and restore a state of equilibrium. Henry Le Chatelier

48. When you take something away from a system at equilibrium, the system shifts in such a way as to replace some what you’ve taken away. Le Chatelier Translated: When you add something to a system at equilibrium, the system shifts in such a way as to use up some of what you’ve added.

49. Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g) Add NH 3 Equilibrium shifts left to offset stress 14.5 The system will shift away from the added component.

50. LeChatelier Example #1 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

51. LeChatelier Example #2 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

52. Le Châtelier’s Principle Changes in Concentration continued ChangeShifts the Equilibrium Increase concentration of product(s)left Decrease concentration of product(s)right Decrease concentration of reactant(s) Increase concentration of reactant(s)right left 14.5 aA + bB cC + dD Add Remove

53. 53 Effects of Changes on the System (continued) Three ways to change pressure: a. Add or remove a gaseous reactant or product. b. Addition of inert gas. (not involved in the reaction) c. Decreasing the volume.

54. 54 N 2 (g) + 3H 2 (g)  2NH 3 (g) Shift toward the side with fewer moles. If you decrease the volume of container the gases decrease their volume to compensate.

55. 55 2NO 2 (g)  N 2 O 4(g) Brown colorless

56. LeChatelier Example #3 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

57. Le Châtelier’s Principle Changes in Volume and Pressure A (g) + B (g) C (g) ChangeShifts the Equilibrium Increase pressureSide with fewest moles of gas Decrease pressureSide with most moles of gas Decrease volume Increase volumeSide with most moles of gas Side with fewest moles of gas 14.5

58. Important note…. Changes in concentration, volume, or pressure will not alter the equilibrium constant.

59. Le Châtelier’s Principle Temperature: K will change depending upon the temperature ChangeExothermic Rx Increase temperatureK decreases Decrease temperatureK increases Endothermic Rx K increases K decreases 14.5 colder hotter (treat the energy change as a reactant or product) N 2 O 4 (g)  2 NO 2 (g)

60. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 60 N 2 O 4 (g)  2 NO 2 (g)

61. LeChatelier Example #4 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

62. uncatalyzedcatalyzed 14.5 Catalyst lowers E a for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium. Adding a Catalyst ??? does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner Le Châtelier’s Principle

63. ChangeShift Equilibrium Change Equilibrium Constant Concentrationyesno Pressureyesno Volumeyesno Temperatureyes Catalystno 14.5