1. Chapter 13 Chemical Equilibrium

2. The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. (DYNAMIC SITUATION)

3. The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate. N 2 O 4 (g) 2 NO 2 (g)

4. A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant.

5. Depicting Equilibrium Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow. N 2 O 4 (g) 2 NO 2 (g)

6. The Equilibrium Constant

7. Consider the generalized reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD

8. Write the equilibrium expression for the following reaction: 4NH 3 (g) + 7O 2 (g) ↔ 4NO 2 (g) + 6H 2 O(g) Equilibrium constants are customarily given without units (page 583 textbook.)

9. Write the equilibrium expression for the following reaction: 4NH 3 (g) + 7O 2 (g) ↔ 4NO 2 (g) + 6H 2 O(g) K = [NO 2 ] 4 [H 2 O] 6 [NH 3 ] 4 [O 2 ] 7

10. The following equilibrium concentrations were observed for the Haber process, N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g), 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 Calculate the value of K at 127°C – 3.8 x 10 4 Calculate the value of K for the reverse reaction. – 2.6 x 10 -5

11. Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. K c = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = 4.72 at 100  C [N 2 O 4 ] [NO 2 ] 2 N2O4 (g)N2O4 (g) 2 NO 2 (g) ↔ ↔

12. Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. K c = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] N 2 O 4(g) 2 NO 2(g) K c = = (0.212) 2 at 100  C [NO 2 ] 4 [N 2 O 4 ] 2 2 N 2 O 4(g) 4 NO 2(g) ↔ ↔

13. Manipulating Equilibrium Constants The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.

14. IMPORTANT The equilibrium expression for a reverse reaction is the reciprocal of the expression for the forward reaction The equilibrium constants for the forward and reverse reactions are therefore also reciprocals of each other. When the balanced equation for a reaction is multiplied by a factor, n, the equilibrium constant for the new reaction is the original constant raised to the power of n. K new = (K original ) n

15. Equilibrium Expressions Involving Pressures Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C c ) (P D d ) (P A a ) (P B b )

16. Relationship Between K c and K p where K p = K c (RT)  n  n = (moles of gaseous product) - (moles of gaseous reactant) (on reference table)

17. 2NO(g) + Cl 2 (g) ↔ 2NOCl(g) The pressures at equilibrium at 25°C were: – P NOCl =1.2 atm P NO = 5.0 x 10 -2 atm – P Cl 2 = 3.0 x 10 -1 atm Calculate Kp – 1.9 x 10 3 Calculate Kc – 4.6 x 10 4

18. Heterogeneous Equilibrium

19. The Concentrations of Solids and Liquids Are Essentially Constant Both can be obtained by multiplying the density of the substance by its molar mass — and both of these are constants at constant temperature.

20. The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression. K c = [Pb 2+ ] [Cl - ] 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq) ↔

21. As long as some CaCO 3 or CaO remain in the system, the amount of CO 2 above the solid will remain the same. CaCO 3 (s) CO 2 (g) + CaO (s) ↔ K c = [CO 2 ]

22. Write the expressions for K and Kp for the following processes: Solid phosphorous pentachloride decomposes to liquid phosphorous trichloride and chlorine gas. – K = [Cl 2 ] Kp = P Cl 2 Deep blue solid copper (II) sulfate pentahydrate is heated to drive off water vapor to form white solid copper (II) sulfate. – K = [H 2 O] 5 Kp = (P H 2 O ) 5

23. What Does the Value of K Mean? If K>>1, the reaction is product-favored; product predominates at equilibrium.

24. What Does the Value of K Mean? If K>>1, the reaction is product-favored; product predominates at equilibrium. If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium.

25. Equilibrium Calculations

26. An Equilibrium Problem A closed system initially containing 1.000 x 10 -3 M H 2 and 2.000 x 10 -3 M I 2 at 448  C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 -3 M. Calculate K c at 448  C for the reaction taking place, which is H 2 (g) + I 2 (s) 2 HI (g) ↔

27. What Do We Know? [H 2 ], M[I 2 ], M[HI], M Initial1.000 x 10 -3 2.000 x 10 -3 0 Change Equilibrium1.87 x 10 -3

28. [HI] Increases by 1.87 x 10 -3 M [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change+1.87 x 10 -3 At equilibrium1.87 x 10 -3

29. Stoichiometry tells us [H 2 ] and [I 2 ] decrease by half as much. [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium1.87 x 10 -3 H 2 (g) + I 2 (s) 2 HI (g) ↔

30. We can now calculate the equilibrium concentrations of all three compounds… [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium6.5 x 10 -5 1.065 x 10 -3 1.87 x 10 -3

31. …and, therefore, the equilibrium constant. K c = [HI] 2 [H 2 ] = 51 = (1.87 x 10 -3 ) 2 (6.5 x 10 -5 )(1.065 x 10 -3 )

32. The Reaction Quotient (Q) (lets us calculate the direction that the reaction must shift to reach equilibrium) Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. To calculate Q, one substitutes the initial concentrations of reactants and products into the equilibrium expression.

33. If Q = K, the system is at equilibrium.

34. If Q > K, there is too much product, and the equilibrium shifts to the left.

35. If Q < K, there is too much reactant, and the equilibrium shifts to the right.

36. K = 6.0 x 10 -2 Predict the direction in which the system will shift to reach equilibrium: [NH 3 ] o =1.0 x 10 -3 M [N 2 ] o =1.0 x 10 -5 M [H 2 ] o =2.0 x 10 -3 M – Q = 1.3 x 10 7 (left) [NH 3 ] o = 2.0 x 10 -4 M [N 2 ] o =1.5 x 10 -5 M [H 2 ] o =3.54 x 10 -1 M – Q = 6.01 x 10 -2 (equilibrium) [NH 3 ] o =1.0 x 10 -4 M [N 2 ] o =5.0 M [H 2 ] o =1.0 x 10 -2 M – Q = 2.0 x 10 -3 (right)

37. N 2 O 4 (g) 2NO 2 (g) N 2 O 4 (g) was placed into a flask and allowed to reach equilibrium at a temperature where Kp = 0.133. At equilibrium, the pressure of N 2 O 4 was 2.71 atm. Calculate the equilibrium pressure of NO 2 (g). – 0.600 atm ↔

38. At a certain temperature a 1.00 L flask initially contained 0.298 mol PCl 3 (g) and 8.70 x 10 -3 mol PCl 5 (g). After the system reached equilibrium, 2.00 x 10 -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. [Cl 2 ] = 2.00 x 10 -3 M [PCl 3 ] = 0.300M [PCl 5 ] = 6.70 x 10 -3 M K = 8.96x10 -2 ↔

39. Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700K the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.000 mol of each component is mixed in a 1.000 L flask. You must calculate Q in order to determine the direction the reaction must proceed to reach equilibrium.

40. The reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15 x 102 at a certain temperature. In a particular experiment, 3.000 mol of each component was added to a 1.500 L flask. Calculate the equilibrium concentrations of all species.

41. SOLVING EQUILIBRIUM PROBLEMS Write the balanced equation Write the equilibrium expression List the initial concentrations Calculate Q and determine the direction of the shift to equilibrium Define the change needed and the equilibrium concentrations Substitute the equilibrium concentrations into the equilibrium expression Check your calculated equilibrium concentrations to make sure they give the correct value of K

42. Oh oh….time to take it up a notch!! Suppose 3.000 mol H 2 and 6.000 mol F 2 are mixed in a 3.000 L flask. Assume that the equilibrium constant for the synthesis of hydrogen fluoride gas at this temperature is 1.15 x 10 2. Calculate the equilibrium concentrations of each of the components. Do you need to calculate Q?

43. Gaseous hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where the equilibrium constant is 1.00 x 10 2. Suppose HI at 5.000 x 10 -1 atm, H 2 at 1.000x 10 -2 atm, and I 2 at 5.000 x 10 -3 atm are mixed in a 5.000L flask. Calculate the equilibrium pressures of all species.

44. If K is small, we’re in luck! If K is small that means that the reaction does not proceed very far to the right. This means that the change, x, in this case will be very small and complicated terms in equilibrium calculations can be eliminated. Let’s see how this may work…..

45. 2NOCl(g) 2NO(g) + Cl 2 (g) 1.0 mol NOCl is placed in a 2.0 L flask K=1.6 x 10-5 What are the equilibrium concentrations? ↔

46. Le Châtelier’s Principle

47. “If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”

48. The Haber Process The transformation of nitrogen and hydrogen into ammonia (NH 3 ) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance. N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)

49. The Haber Process If H 2 is added to the system, N 2 will be consumed and the two reagents will form more NH 3.

50. The Haber Process This apparatus helps push the equilibrium to the right by removing the ammonia (NH 3 ) from the system as a liquid.

51. As 4 O 6 (s) + 6C(s) ↔ As 4 (g) + 6CO(g) Predict the direction of the shift: Addition of CO Addition of C Removal of tetraarsenic hexoxide Removal of gaseous arsenic Addition of tetraarsenic hexoxide

52. The Haber Process N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) + 92 kJ How will a change in temperature affect this reaction? Only changes in temperature can change K!

53. For each of the following reactions, predict how the value of K changes as the temperature is increased. N 2 (g) + O 2 (g) ↔ 2NO(g) ΔH° = 181 kJ 2SO 2 (g) + O 2 (g) ↔ 2SO 3 (g) ΔH° = -198 kJ

54. The Effect of Changes in Pressure Pressure can be changed by: – Adding or removing a gaseous reactant or product – Adding an inert gas (has no effect on the equilibrium position) – Changing the volume of the container

55. Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced. P 4 (s) + 6Cl 2 (g) ↔ 4PCl 3 (l) PCl 3 (g) + Cl 2 (g) ↔ PCl 5 (g) PCl 3 (g) + 3NH 3 (g) ↔ P(NH 2 ) 3 (g) + 3HCl(g)

56. Catalysts

57. Catalysts increase the rate of both the forward and reverse reactions.

58. Catalysts When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered.