Chapter 12 Chemical Equilibrium

1. Chapter 12Chemical Equilibrium

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

3. The Concept of Equilibrium • Consider colorless frozen N2O4. At room temperature, it decomposes to brown NO2: • N2O4(g)  2NO2(g). • At some time, the color stops changing and we have a mixture of N2O4 and NO2. • Chemical equilibrium is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. At that point, the concentrations of all species are constant. • Using the collision model: • as the amount of NO2 builds up, there is a chance that two NO2 molecules will collide to form N2O4. • At the beginning of the reaction, there is no NO2 so the reverse reaction (2NO2(g)  N2O4(g)) does not occur.

4. The Concept of Equilibrium • As the substance warms it begins to decompose: • N2O4(g)  2NO2(g) • When enough NO2 is formed, it can react to form N2O4: • 2NO2(g)  N2O4(g). • At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4 • The double arrow implies the process is dynamic.

5. 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.

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

7. N2O4 (g) 2 NO2 (g) Depicting Equilibrium In a system at equilibrium, both the forward and reverse reactions are being carried out; as a result, we write its equation with a double arrow

8. The Equilibrium Constant

9. The Equilibrium Constant • Forward reaction: N2O4 (g) 2 NO2 (g) • Rate law: Rate = kf [N2O4]

10. The Equilibrium Constant • Reverse reaction: 2 NO2 (g) N2O4 (g) • Rate law: Rate = kr [NO2]2

11. [NO2]2 [N2O4] kf kr = The Equilibrium Constant • Therefore, at equilibrium Ratef = Rater kf [N2O4] = kr [NO2]2 • Rewriting this, it becomes

12. [NO2]2 [N2O4] Keq = kf kr = The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes

13. aA + bB cC + dD [C]c[D]d [A]a[B]b Kc = The Equilibrium Constant • To generalize this expression, consider the reaction • The equilibrium expression for this reaction would be

14. The Equilibrium Constant • Kc is based on the molarities of reactants and products at equilibrium. • We generally omit the units of the equilibrium constant. • Note that the equilibrium constant expression has products over reactants.

15. The Equilibrium Expression • Write the equilibrium expression for the following reaction:

16. What Are the Equilibrium Expressions for These Equilibria?

17. (PC)c (PD)d (PA)a (PB)b Kp = The Equilibrium Constant Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written

18. n V P = RT Relationship between Kc and Kp • From the ideal gas law we know that PV = nRT • Rearranging it, we get

19. Relationship between Kc and Kp Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes Kp= Kc(RT)n Where n = (moles of gaseous product) − (moles of gaseous reactant)

20. N2O4 (g) 2 NO2 (g) Equilibrium Can Be Reached from Either Direction As you can see, the ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.

21. Equilibrium Can Be Reached from Either Direction This is the data from the last two trials from the table on the previous slide.

22. Equilibrium Can Be Reached from Either Direction It does not matter whether we start with N2 and H2 or whether we start with NH3. We will have the same proportions of all three substances at equilibrium.

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. Kc = = 4.72 at 100C Kc = = 0.212 at 100C N2O4(g) 2 NO2(g) 2 NO2(g) N2O4(g) [NO2]2 [N2O4] [N2O4] [NO2]2 1 0.212 = Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

26. Kc = = 0.212 at 100C Kc = = (0.212)2 at 100C N2O4(g) 2 NO2(g) 2 N2O4(g) 4 NO2(g) [NO2]4 [N2O4]2 [NO2]2 [N2O4] 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.

27. 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.

28. Heterogeneous Equilibrium

29. The Equilibrium Constant • Heterogeneous Equilibria • When all reactants and products are in one phase, the equilibrium is homogeneous. • If one or more reactants or products are in a different phase, the equilibrium is heterogeneous. • Consider: • experimentally, the amount of CO2 does not seem to depend on the amounts of CaO and CaCO3. Why?

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

31. The Equilibrium Constant • Heterogeneous Equilibria • Neither density nor molar mass is a variable, the concentrations of solids and pure liquids are constant. (You can’t find the concentration of something that isn’t a solution!) • We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions. • The amount of CO2 formed will not depend greatly on the amounts of CaO and CaCO3 present. • Kc = [CO2]

32. CaCO3 (s) CO2 (g) + CaO(s) As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.

33. PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq) The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression Kc = [Pb2+] [Cl−]2

34. Equilibrium Calculations

35. Calculating Equilibrium Constants • Steps to Solving Problems: • Write an equilibrium expression for the balanced reaction. • Write an ICE table. Fill in the given amounts. • Use stoichiometry (mole ratios) on the change in concentration line. • Deduce the equilibrium concentrations of all species. • Usually, the initial concentration of products is zero. (This is not always the case.)

36. H2 (g) + I2 (g) 2 HI (g) Equilibrium Calculations A closed system initially containing 1.000 x 10−3 M H2 and 2.000 x 10−3 M I2 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 Kc at 448C for the reaction taking place, which is

37. H2 (g) + I2 (g) 2 HI (g) What Do We Know?

38. H2 (g) + I2 (g) 2 HI (g) [HI] Increases by 1.87 x 10-3M

39. H2 (g) + I2 (g) 2 HI (g) Stoichiometry tells us [H2] and [I2]decrease by half as much

40. H2 (g) + I2 (g) 2 HI (g) We can now calculate the equilibrium concentrations of all three compounds…

41. Kc= [HI]2 [H2] [I2] (1.87 x 10-3)2 (6.5 x 10-5)(1.065 x 10-3) = = 51 …and, therefore, the equilibrium constant

42. Applications of Equilibrium Constants • Predicting the Direction of Reaction • We define Q, the reaction quotient, for a reaction at conditions NOT at equilibrium • as • where [A], [B], [P], and [Q] are molarities at any time. • Q = K only at equilibrium.

43. The Reaction Quotient (Q) • To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. • Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

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

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

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

47. Applications of Equilibrium Constants • Predicting the Direction of Reaction • If Q > K then the reverse reaction must occur to reach equilibrium (go left) • If Q < K then the forward reaction must occur to reach equilibrium (go right)

48. If Q > Keq, shift to left (toward reactant) If Q < Keq, shift to right (toward product)

49. Le Châtelier’s Principle

50. 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.”