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

2. Equilibrium The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. Not in equal amounts!! Their rates are the same!!! © 2012 Pearson Education, Inc.

3. Equilibrium A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant. Nothing can escape the system for equil to occur. © 2012 Pearson Education, Inc.

4. 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.

5. Equilibrium Equilibrium  Equal The rates of the forward and reverse reactions are equal at equilibrium. But that does not mean the concentrations of reactants and products are equal. Some reactions reach equilibrium only after almost all the reactant molecules are consumed; we say the position of equilibrium favors the products. Other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed; we say the position of equilibrium favors the reactants.

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. -Products over reactants -The ratio of the rate constants is a constant at that temperature

10. 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. K = equilibrium constant, NO UNITS! BOARD

11. Equilibrium Writing Equilibrium-Constant Expressions Write the equilibrium expression for K c for the following reactions:

12. Equilibrium The Equilibrium Constant 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.

13. Equilibrium Relationship Between K c and K p From the ideal-gas law we know that Rearranging it, we get PV = nRT P = RT nVnV © 2012 Pearson Education, Inc.

14. Equilibrium Relationship Between K c and K p Plugging this into the expression for K p, for each substance, the relationship between K c and K p becomes K p = K c (RT)  n  n = (moles of gaseous product)  (moles of gaseous reactant) © 2012 Pearson Education, Inc.

15. Equilibrium Converting between K c and K p For the process, K c = 9.60 at 300  C. Calculate K p for this reaction at this temperature. Solution (1 N 2 + 3 H 2 ),  n = 2 – 4 =-2 273 + 300 = 573 K

16. Equilibrium 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. © 2012 Pearson Education, Inc.

17. 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. - If K is approx. equal to 1, reactants and products are equally favored

18. 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.

19. 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.

20. 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. © 2012 Pearson Education, Inc.

21. Equilibrium Combining Equilibrium Expressions Given the reactions determine the value of K c for the reaction X2 reverse target squared inverse

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

23. 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.

24. Equilibrium The Concentrations of Solids and Liquids Are Essentially Constant 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.

25. 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.

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

27. 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.

28. 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 Equilibrium1.87 x 10  3 © 2012 Pearson Education, Inc.

29. 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.

30. 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.

31. 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.

32. 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.

33. Equilibrium STEPS to Calculating K’s 1.If eq conc are given, plug them into Kc or Kp and solve. 2.Tabulate all known initial and eq conc of the species that appear in the eq expression and start the ICE BOX. 3.For initial conc, calculate the change as the system reaches equil. 4.Use the stoich coefficients to calc all other changes 5.Calculate eq concentrations, add/subtract 6.Plug final conc into Keq and solve. Quadratic?

34. 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, substitute the initial concentrations of reactants and products into the equilibrium expression and solve. © 2012 Pearson Education, Inc.

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

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

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

38. Equilibrium Predicting the Direction of Approach to Equilibrium The initial concentrations are The reaction quotient is therefore 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. Because Q c < K c, the concentration of HI must increase and the concentrations of H 2 and I 2 must decrease to reach equilibrium; the reaction as written proceeds left to right to attain equilibrium.

39. Equilibrium 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?

40. Equilibrium Calculating Equilibrium Concentration from Initial Concentrations A 1.000-L flask is filled with 1.000 mol of H 2 (g) and 2.000 mol of I 2 (g) at 447  C. The value of the equilibrium constant K c for the reaction at 448  C is 50.5.What are the equilibrium concentrations of H 2, I 2, and HI in moles per liter? [H 2 ] = 1.000 M and [I 2 ] = 2.000 M

41. Equilibrium Calculating Equilibrium Concentration from Initial Concentrations

42. Equilibrium Calculating Equilibrium Concentration from Initial Concentrations Comment Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions to the equation will give you a value that leads to negative concentrations and thus is not chemically meaningful. Reject this solution to the quadratic equation.

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.

45. Equilibrium Le Châtelier’s Principle The Rules: Add a reactant, shift to the products. Add a product, shift to the reactants. Increase temperature, shift away from heat. Decrease temperature, shift toward heat. Increase pressure (reduce volume), shift toward side with fewer moles of gas. Decrease pressure (increase volume), shift toward side with more moles of gas. Concentration of solids and liquids cannot change, adding a solid or liquid will not shift the equilibrium. The only stress that changes the K eq is temperature. A catalyst does not change the equilibrium… it allows the system to reach the same equilibrium more quickly.

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. Equilibrium Changes in Temperature Co(H 2 O) 6 2+ (aq) + 4Cl(aq)CoCl 4 (aq) + 6H 2 O(l) © 2012 Pearson Education, Inc. This is an endothermic rxn, which side of the equation does the heat belong? PINK BLUE

50. Equilibrium Catalysts © 2012 Pearson Education, Inc.

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

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