1. Chapter 15 Chemical Equilibrium © 2009, Prentice-Hall, Inc.

2. The Concept of Equilibrium © 2009, Prentice-Hall, Inc.

3. The Concept of Equilibrium © 2009, Prentice-Hall, Inc.

4. A System at Equilibrium © 2009, Prentice-Hall, Inc.

5. The Equilibrium Constant Forward reaction: N 2 O 4 (g)  2 NO 2 (g) Rate Law: Rate = k f [N 2 O 4 ] Reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) Rate Law: Rate = k r [NO 2 ] 2

6. The Equilibrium Constant Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] = K eq = kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =

7. The Equilibrium Constant © 2009, Prentice-Hall, Inc. K c = [C] c [D] d [A] a [B] b aA + bBcC + dD

8. 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 ) K p = K c (RT)  n  n = (moles of gaseous product) - (moles of gaseous reactant)

9. Equilibrium Can Be Reached from Either Direction © 2009, Prentice-Hall, Inc.

10. Equilibrium Can Be Reached from Either Direction © 2009, Prentice-Hall, Inc.

11. Equilibrium Can Be Reached from Either Direction It doesn’t matter whether we start with N 2 and H 2 or whether we start with NH 3 : we will have the same proportions of all three substances at equilibrium. © 2009, Prentice-Hall, Inc.

12. What Does the Value of K Mean? If K>>1, the reaction is product-favored; product predominates at equilibrium. © 2009, Prentice-Hall, Inc.

13. What Does the Value of K Mean? If K>>1, the reaction is product-favored; product predominates at equilibrium. © 2009, Prentice-Hall, Inc. If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium.

14. Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. © 2009, Prentice-Hall, Inc. 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)

15. 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. © 2009, Prentice-Hall, Inc. 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)

16. The Concentrations of Solids and Liquids Are Essentially Constant The concentrations of solids and liquids do not appear in the equilibrium expression. © 2009, Prentice-Hall, Inc. K c = [Pb 2+ ] [Cl - ] 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq)

17. As long as some CaCO 3 or CaO remain in the system, the amount of CO 2 above the solid will remain the same. © 2009, Prentice-Hall, Inc. CaCO 3 (s) CO 2 (g) + CaO (s)

18. 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 © 2009, Prentice-Hall, Inc. H 2 (g) + I 2 (g) 2 HI (g)

19. 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 © 2009, Prentice-Hall, Inc.

20. [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 © 2009, Prentice-Hall, Inc.

21. 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 © 2009, Prentice-Hall, Inc.

22. 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 © 2009, Prentice-Hall, Inc.

23. …and, therefore, the equilibrium constant. © 2009, Prentice-Hall, Inc. 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 )

24. 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 of reactants and products into the equilibrium expression. © 2009, Prentice-Hall, Inc.

25. If Q = K, © 2009, Prentice-Hall, Inc. the system is at equilibrium.

26. If Q > K, © 2009, Prentice-Hall, Inc. there is too much product, and the equilibrium shifts to the left.

27. If Q < K, © 2009, Prentice-Hall, Inc. there is too much reactant, and the equilibrium shifts to the right.

28. 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.” © 2009, Prentice-Hall, Inc.

29. 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. © 2009, Prentice-Hall, Inc.

30. 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. © 2009, Prentice-Hall, Inc.

31. The Effect of Changes in Temperature

32. Catalysts Catalysts increase the rate of both the forward and reverse reactions. © 2009, Prentice-Hall, Inc.

33. Catalysts When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered. © 2009, Prentice-Hall, Inc.