3. Get two 10 mL graduated cylinders Place 5 mL in one, and 10 mL in the other Place a straw in each cylinder and transfer water from one tube to the other

4. Make a graph of Volume vs # of transfers Volume mL # of transfers

5. If a volatile liquid is placed in an open container underneath an enclosed space the liquid will evaporate until the air in the closed container is saturated with particles of the vapour.

6. When the rate of evaporation is equal to the rate of condensation, no net change is occurring. The system is in a state of dynamic equilibrium since both events occur simultaneously.

7. Most elementary reactions are, in theory, reversible so they are capable of establishing a dynamic equilibrium. The gas, NO 2 is composed of small rapidly moving particles which collide randomly. Sometimes these collisions result in the formation of N 2 O 4 molecules which are capable of spontaneously reforming NO 2 molecules after breaking apart. This gas is really a mixture of 2 gases where the rates of formation of each from the other is equal when in dynamic equilibrium.

8. NO2NO2 N2O4N2O4 NO2 -> N2O4 Movie

9. This dynamic equilibrium is represented by the following equation: NO 2 (g) + NO 2 (g) N 2 O 4 (g) 2NO 2 (g) N 2 O 4 (g) or

10. A system at equilibrium can be disturbed in a number of ways. When disturbed it tends to respond in a way which offsets the disturbance. This is called LeChatelier’s Principle. To illustrate this consider the following system.

11. If water is added to the left water flows from left to right

12. If water is removed from the left water flows from right to left

13. This principle can be applied to an equilibrium. Consider this system A(g) + B (g) C(g) + D(g) If some A is added to the system the forward reaction will go faster, using up A and B faster than it can be replaced by C and D reacting in the reverse reaction. The net result can be shown by the following graphs.

14. A(g) + B (g) C(g) + D(g) [A][A] [B][B] time [C][C] [D][D] Some A is added New Equilibrium is established

15. A(g) + B (g) C(g) + D(g) [A][A] [B][B] time [C][C] [D][D] Some B is added New Equilibrium is established

16. A(g) + B (g) C(g) + D(g) [A][A] [B][B] time [C][C] [D][D] New Equilibrium is established Some C is added

17. A(g) + B (g) C(g) + D(g) [A][A] [B][B] time [C][C] [D][D] New Equilibrium is established Some D is added

18. A(g) + B (g) C(g) + D(g) [A][A] [B][B] time [C][C] [D][D] Some A is removed New Equilibrium is established

19. A(g) + B (g) C(g) + D(g) [A][A] [B][B] time [C][C] [D][D] Some B is removed New Equilibrium is established

20. A(g) + B (g) C(g) + D(g) [A][A] [B][B] time [C][C] [D][D] New Equilibrium is established Some C is removed

21. A(g) + B (g) C(g) + D(g) [A][A] [B][B] time [C][C] [D][D] New Equilibrium is established Some D is removed

22. 2A(g) + B (g) 4C(g) + 3D(g) [A][A] [B][B] time [C][C] [D][D] Some A is added New Equilibrium is established 2x x 4x3x

23. 4A (g) + 2B (g) 3C (g) + D (g) [A][A] [B][B] time [C][C] [D][D] Some B is removed New Equilibrium is established 2x x 4x 3x

24. 4A (g) + 2B (g) 3C (g) + D (g) [B][B] time [C][C] [D][D] Some B is removed 2x x 3x [A][A] 4x or shown all on 1 graph

25. How do changes in Temperature affect a system at equilibrium? To consider this let’s look at this equilibrium. 2NO 2 (g) N 2 O 4 (g) When this equilibrium is upset by heating, it becomes darker brown. Does this suggest this reaction is endothermic or exothermic? Hint: Increasing temperature is just like adding heat. Brown Colourless

26. 2NO 2 (g) N 2 O 4 (g)+ heat If an equilibrium is cooled, the equilibrium shifts to the side with the heat as the equilibrium responds to this disturbance by replacing the heat which has been removed. Graphs for the system above are:

27. [NO 2 ] time 2x 2NO 2 (g) N 2 O 4 (g)+ heat The system is plunged into ice water(removal of heat) [N2O4][N2O4] time x

28. 2A(g)+B (g) + heat 4C(g) + 3D(g) [A][A] [B][B] time [C][C] [D][D] The temperature is increased New Equilibrium is established 2x x 4x3x

29. How do changes in Volume affect a system at equilibrium? To consider this let’s look at this equilibrium.Volume 2NO 2 (g) N 2 O 4 (g)

30. When the volume is decreased what happens?

31. 2NO 2 (g) N 2 O 4 (g)

32. When the volume is decreased what happens? 2NO 2 (g) N 2 O 4 (g) The gas appears browner since it is more concentrated

33. When the volume is decreased what happens? 2NO 2 (g) N 2 O 4 (g) Then it gets lighter, due to the NO 2 changing into N 2 O 4

34. When the volume is decreased what happens? 2NO 2 (g) N 2 O 4 (g) Then it gets lighter, due to the NO 2 changing into N 2 O 4

35. When the volume is decreased what happens? 2NO 2 (g) N 2 O 4 (g) Then it gets lighter, due to the NO 2 changing into N 2 O 4

36. When the volume is decreased what happens? 2NO 2 (g) N 2 O 4 (g) This happens because one N 2 O 4 molecule takes up less space than 2 NO 2 molecules

37. 2NO 2 (g) N 2 O 4 (g) When this equilibrium is upset by a decrease in volume, it shifts in a direction which decreases the total number of molecules. Since there is less space it makes sense to reduce the number of molecules present. If the volume is doubled ….

38. 2NO 2 (g) N 2 O 4 (g) When the volume is doubled ……

39. 2NO 2 (g) N 2 O 4 (g) When the volume is doubled ……

40. 2NO 2 (g) N 2 O 4 (g) When the volume is doubled ……

41. 2NO 2 (g) N 2 O 4 (g) It first gets lighter since the NO 2 molecules are less concentrated

42. 2NO 2 (g) N 2 O 4 (g) Then it gets darker because with more space available the N 2 O 4 molecules can form twice as many NO 2 molecules

43. 2NO 2 (g) N 2 O 4 (g) Then it gets darker because with more space available the N 2 O 4 molecules can form twice as many NO 2 molecules

44. 2NO 2 (g) N 2 O 4 (g) Then it gets darker because with more space available the N 2 O 4 molecules can form twice as many NO 2 molecules

45. 2NO 2 (g) N 2 O 4 (g) Then it gets darker because with more space available the N 2 O 4 molecules can form twice as many NO 2 molecules

46. [NO 2 ] time 2x 2NO 2 (g) N 2 O 4 (g)+ heat The volume is doubled. [N2O4][N2O4] time x

47. 2NO 2 (g) N 2 O 4 (g)+ heat This shift in equilibrium can be demonstrated mathematically using rate equations What are the rate equations for forward and reverse reactions? r f = k[NO 2 ] 2, r r = k[N 2 O 4 ] if concentrations of NO 2 and N 2 O 4 are 1 mol/L r f = k[1] 2 = 1kr r = k[1] = 1k

48. 2NO 2 (g) N 2 O 4 (g)+ heat This shift in equilibrium can be demonstrated mathematically using rate equations What are the rate equations for forward and reverse reactions? r f = k[NO 2 ] 2, r r = k[N 2 O 4 ] if concentrations of NO 2 and N 2 O 4 are 1 mol/0.5L r f = k[2] 2 = 4kr r = k[2] = 2k How does this impact on Equilibrium?

49. 2NO 2 (g) N 2 O 4 (g)+ heat What happens when the volume is changed from 1 L to 2 L? Assume 1 mol of each gas. r f = k[NO 2 ] 2, r r = k[N 2 O 4 ] [NO 2 ]and [N 2 O 4 ]are 1 mol/2L = 0.5 mol/L r f = k[0.5] 2 = 0.25k, r r = k[0.5] = 0.5k How does this impact on Equilibrium?

50. 2NO 2 (g) N 2 O 4 (g)+ heat r f = k[0.5] 2 = 0.25k, r r = k[0.5] = 0.5k Graph Rate change vs time r f =r r

51. 2NO 2 (g) N 2 O 4 (g)+ heat Describe the graph above r f =r r

52. 2NO 2 (g) N 2 O 4 (g)+ heat The increase in volume leads to decreased probability of collisions so both rates drop, r f drops twice as much, since r f = 0.25k, r r = 0.5k. More N 2 O 4 is used than is made so r r slows down. More NO 2 is made than is used so r f increases until both rates are equal and a new equilibrium is established. r f =r r

53. Under a specific set of conditions of temperature and pressure all participants in a dynamic equilibrium are related mathematically. To determine this mathematical relationship the following data will be considered for this equilibrium: Cl 2 (g) + PCl 3 (g) PCl 5 (g)

54. Hint: Try to find a mathematical constant [PCl 5 ] [PCl 3 ][Cl 2 ] = 0.800

55. This means for any equilibrium aA + bB cC + dD Ke = [C] c [D] d [A] a [ B] b Find the equilibrium constant for this reaction: H 2 (g) + I 2 (g) 2HI(g) If [H 2 ](g) = 0.25 molL -1, [I 2 ](g)= 0.45 molL -1, [HI](g)= 0.0035 molL -1

56. Ke = [HI] 2 [H 2 ] 1 [ I 2 ] 1 Ke = [0.0035] 2 [0.25] 1 [ 0.45] 1 Ke = 1.1 x 10 -4

57. For this equilibrium 2A + B C + D If Ke is very large there is a lot of C and D and very little of A and B (product favoured reaction) If Ke is very small there is a lot of A and B and very little of C and D (reactant favoured reaction)

58. If an equilibrium involves more than one phase (solids and aqueous ions for instance) the equilibrium expression only considers freely mobile particles since they alone obey the Laws of Mass Action. For example the Ke expression for this equilibrium PbI 2 (s) Pb 2+ (aq) + 2I 1- (aq) is Ke = [Pb 2+ ][I 1- ] 2 since the solid PbI 2 is not composed of freely mobile particles

59. 1.For the equilibrium PCl 3 (g) + Cl 2 (g) PCl 5 (g) 0.50 mol of PCl 5 (g) is placed in a 5.0 L container. At equilibrium 0.14 mol of Cl 2 (g) is present. Calculate the equilibrium concentrations of each component and find Ke.

60. PCl 3 (g) + Cl 2 (g) PCl 5 (g) (molL -1 ) Initial shift @E O.5 mol 5.0 L = 0.10 molL -1 0.028 Where did the 0.028 molL -1 of the Cl 2 come from? It came from the reaction of 0.028 molL -1 of the PCl 5. When the Cl 2 was created an equal amount of PCl 3 was made The amount of PCl 5 at equilibrium can now be determined by subtraction. 0.028 +0.028 -0.028 0.072 Ke can now be calculated. Ke = 0.072 / (0.028)(0.028) = 92

61. 2.0For the equilibrium CO(g) + 2H 2 (g) CH 3 OH(g) 1.60 mol of CH 3 OH(g) is placed in a 4.0 L container. At equilibrium 0.24 mol of CO(g) is present. Calculate the equilibrium concentrations of each component and find Ke.

62. CO(g) + 2H 2 (g) CH 3 OH(g) (molL -1 ) Initial shift @E 1.60 mol 4.0 L = 0.40 molL -1 0.12 Where did the 0.060 molL -1 of the CO come from? It came from the reaction of 0.060 molL -1 of the CH 3 OH. When the CO was created, double the amount of H 2 was made. The amount of CH 3 OH at equilibrium can now be determined by subtraction. 0.060 2x 0.060 +0.060 -0.060 0.34 Ke can now be calculated. Ke = 0.34 / (0.060)(0.12) 2 = 3.9 x 10 2 mol -2 L 2

63. 3.0For the equilibrium I 2 (g) + H 2 (g) 2HI(g) H 2 (g) and I 2 (g) were placed in a container at 728 K and allowed to reach equilibrium (eq.) At eq. the [H 2 ] = 0.23 M, [ I 2 ] = 0.28 M, and [HI] = 3.4 M. If in another 2.0 L container at the same temperature 0.64 mol of H 2 and 0.64 mol of I 2 are injected. Find the equilibrium concentrations of all 3 gases.

64. I 2 (g) + H 2 (g) 2HI(g) Since all the concentrations are given for a specific temperature find the Ke at this temperature. Ke = [HI] 2 [I 2 ] [H 2 ] = [3.4] 2 [0.28] [0.23] Ke = 179.5, this value can now be used to answer the rest of the question

65. I 2 (g) + H 2 (g) 2HI(g) (molL -1 ) Initial shift @E 0.64 mol 2.0 L = 0.32 molL -1 Since there is no HI in the container initially, the system, in order to reach equilibrium, must produce some. To do this it must use up equal quantities of H 2 and I 2. Since this quantity is unknown let it be x. -x +2x 0.32 The balanced equation shows twice as much HI is produced. Equilibrium values can now be calculated. 0.32 - x 2x

66. Since Ke has already been calculated (179.5) the value of x can be determined using the equilibrium expression. 179.5 = (2x) 2 / (0.32- x) 2 To avoid a quadratic equation take the square root of each side.13.40 = 2x / 0.32 - x 4.287 - 13.40 x = 2x x = 0.278 so [I 2 ] = [H 2 ] = 0.04 M, [HI] = 0.56 M I 2 (g) + H 2 (g) 2HI(g) (molL -1 ) Initial shift @E 0.64 mol 2.0 L = 0.32 molL -1 -x +2x 0.32 0.32 - x 2x