1. Chapter 15 Chemical Equilibrium Dr. Subhash Goel South GA State College Douglas, GA Lecture Presentation © 2012 Pearson Education, Inc.

2. Equilibrium The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. © 2012 Pearson Education, Inc.

3. Equilibrium 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. © 2012 Pearson Education, Inc.

4. Equilibrium A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant. © 2012 Pearson Education, Inc.

5. Equilibrium 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: N2O4(g)N2O4(g)2NO 2 (g) © 2012 Pearson Education, Inc.

6. Equilibrium The Equilibrium Constant © 2012 Pearson Education, Inc.

7. Equilibrium The Equilibrium Constant Forward reaction: N 2 O 4 (g)  2NO 2 (g) Rate law: Rate = k f [N 2 O 4 ] © 2012 Pearson Education, Inc.

8. Equilibrium The Equilibrium Constant Reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) Rate law: Rate = k r [NO 2 ] 2 © 2012 Pearson Education, Inc.

9. Equilibrium The Equilibrium Constant Therefore, at equilibrium Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 Rewriting this, it becomes kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] = © 2012 Pearson Education, Inc.

10. Equilibrium The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes K eq = kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] = © 2012 Pearson Education, Inc.

11. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Exercise 1 Writing Equilibrium-Constant Expressions Write the equilibrium expression for K c for the following reactions: (a) (b) (c)

12. Equilibrium The Equilibrium Constant 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 © 2012 Pearson Education, Inc.

13. Equilibrium The Equilibrium Constant 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 ) © 2012 Pearson Education, Inc.

14. Equilibrium Relationship Between K c and K p The relationship between K c and K p can be express by following equation: where: K p = K c (RT)  n  n = (moles of gaseous product)  (moles of gaseous reactant) © 2012 Pearson Education, Inc.

15. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 2 Converting between K c and K p For the Haber process, K c = 9.60 at 300  C. Calculate K p for this reaction at this temperature.

16. Equilibrium Equilibrium Can Be Reached from Either Direction As you can see, the ratio of [NO 2 ] 2 to [N 2 O 4 ] remains constant at this temperature no matter what the initial concentrations of NO 2 and N 2 O 4 are. © 2012 Pearson Education, Inc.

17. Equilibrium Equilibrium Can Be Reached from Either Direction These are the data from the last two trials from the table on the previous slide. © 2012 Pearson Education, Inc.

18. Equilibrium 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. © 2012 Pearson Education, Inc.

19. Equilibrium 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 ] K c = = 4.72 at 100  C [N 2 O 4 ] [NO 2 ] 2 N2O4(g)N2O4(g)2NO 2 (g) N2O4(g)N2O4(g) © 2012 Pearson Education, Inc.

20. Equilibrium 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 ] K c = = (0.212) 2 at 100  C [NO 2 ] 4 [N 2 O 4 ] 2 4NO 2 (g)2N 2 O 4 (g) N2O4(g)N2O4(g)2NO 2 (g) © 2012 Pearson Education, Inc.

21. Equilibrium 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. (Similar to Hess Law) © 2012 Pearson Education, Inc.

22. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Exercise 3 Combining Equilibrium Expressions Given the reactions determine the value of K c for the reaction

23. Equilibrium Heterogeneous Equilibrium © 2012 Pearson Education, Inc.

24. Equilibrium The Concentrations of Solids and Liquids Are Essentially Constant Both the concentrations of solids and liquids can be obtained by multiplying the density of the substance by its molar mass—and both of these are constants at constant temperature. © 2012 Pearson Education, Inc.

25. Equilibrium 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) + 2Cl  (aq) © 2012 Pearson Education, Inc.

26. Equilibrium 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) © 2012 Pearson Education, Inc.

27. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Exercise 4 Writing Equilibrium-Constant Expressions for Heterogeneous Reactions Write the equilibrium-constant expression K c for (a) (b)

28. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Exercise 5 Analyzing a Heterogeneous Equilibrium Solve Equilibrium can be reached in all cases except (c) as long as sufficient quantities of solids are present. (a) CaCO 3 simply decomposes, forming CaO(s) and CO 2 (g) until the equilibrium pressure of CO 2 is attained. There must be enough CaCO 3, however, to allow the CO 2 pressure to reach equilibrium. (b) CO 2 continues to combine with CaO until the partial pressure of the CO 2 decreases to the equilibrium value. (c) There is no CaO present, so equilibrium cannot be attained because there is no way the CO 2 pressure can decrease to its equilibrium value (which would require some of the CO 2 to react with CaO). (d) The situation is essentially the same as in (a): CaCO 3 decomposes until equilibrium is attained. The presence of CaO initially makes no difference. Each of these mixtures was placed in a closed container and allowed to stand: (a)CaCO 3 (s) (b)CaO(s) and CO 2 (g) at a pressure greater than the value of K p (c)CaCO 3 (s) and CO 2 (g) at a pressure greater than the value of K p (d)CaCO 3 (s) and CaO(s) Determine whether or not each mixture can attain the equilibrium

29. Equilibrium Equilibrium Calculations © 2012 Pearson Education, Inc.

30. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Exercise 6 Calculating K When All Equilibrium Concentrations Are Known After a mixture of hydrogen and nitrogen gases in a reaction vessel is allowed to attain equilibrium at 472  C, it is found to contain 7.38 atm H 2, 2.46 atm N 2, and 0.166 atm NH 3. From these data, calculate the equilibrium constant K p for the reaction

31. Equilibrium 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)2HI(g) © 2012 Pearson Education, Inc.

32. Equilibrium What Do We Know? [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10  3 2.000 x 10  3 0 Change At equilibrium1.87 x 10  3 © 2012 Pearson Education, Inc.

33. Equilibrium [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 © 2012 Pearson Education, Inc.

34. Equilibrium 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 © 2012 Pearson Education, Inc.

35. Equilibrium 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 © 2012 Pearson Education, Inc.

36. Equilibrium and, therefore, the equilibrium constant: K c = [HI] 2 [H 2 ] [I 2 ] = 51 = (1.87 x 10  3 ) 2 (6.5 x 10  5 )(1.065 x 10  3 ) © 2012 Pearson Education, Inc.

37. Equilibrium The Reaction Quotient (Q) 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 on reactants and products into the equilibrium expression. © 2012 Pearson Education, Inc.

38. Equilibrium If Q = K, the system is at equilibrium. © 2012 Pearson Education, Inc.

39. Equilibrium If Q > K, there is too much product, and the equilibrium shifts to the left. © 2012 Pearson Education, Inc.

40. Equilibrium If Q < K, there is too much reactant, and the equilibrium shifts to the right. © 2012 Pearson Education, Inc.

41. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Exercise 7 Predicting the Direction of Approach to Equilibrium At 448  C the equilibrium constant K c for the reaction is 50.5. Predict in which direction the reaction proceeds to reach equilibrium if we start with 2.0  10  2 mol of HI, 1.0  10  2 mol of H 2, and 3.0  10  2 of I 2 in a 2.00-L container.

42. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Exercise 8 Calculating Equilibrium Concentrations For the Haber process,, K p = 1.45  10  5 at 500  C. In an equilibrium mixture of the three gases at 500  C, the partial pressure of H 2 is 0.928 atm and that of N 2 is 0.432 atm. What is the partial pressure of NH 3 in this equilibrium mixture?

43. Equilibrium Le Châtelier’s Principle © 2012 Pearson Education, Inc.

44. Equilibrium Le Châtelier’s Principle “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.” © 2012 Pearson Education, Inc.

46. Equilibrium 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. © 2012 Pearson Education, Inc.

47. Equilibrium 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. © 2012 Pearson Education, Inc.

48. Equilibrium The Haber Process This apparatus helps push the equilibrium to the right by removing the ammonia (NH 3 ) from the system as a liquid. © 2012 Pearson Education, Inc.

49. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Exercise 9 Using Le Châtelier’s Principle to Predict Shifts in Equilibrium Consider the equilibrium In which direction will the equilibrium shift when (a) N2O 4 is added, (b) NO 2 is removed, (c) the pressure is increased by addition of N 2 (g), (d) the volume is increased, (e) the temperature is decreased?

50. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Exercise 10 Predicting the Effect of Temperature on K (a)Using the standard heat of formation data in Appendix C, determine the standard enthalpy change for the reaction (b)Determine how the equilibrium constant for this reaction should change with temperature.

51. Equilibrium Catalysts © 2012 Pearson Education, Inc.

52. Equilibrium Catalysts Catalysts increase the rate of both the forward and reverse reactions. © 2012 Pearson Education, Inc.

53. Equilibrium Catalysts When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered. © 2012 Pearson Education, Inc.