Chapter 15 Chemical Equilibrium

1. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 15Chemical Equilibrium John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall

2. 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 concentrations of all species are constant.

3. The point at which the rate of decomposition: • N2O4(g)  2NO2(g) • equals the rate of dimerization: • 2NO2(g)  N2O4(g). • is dynamic equilibrium. • The equilibrium is dynamic because the reaction has not stopped: the opposing rates are equal. • Consider frozen N2O4: only white solid is present. On the microscopic level, only N2O4 molecules are present.

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

5. As the substance warms it begins to decompose: • N2O4(g)  2NO2(g) • A mixture of N2O4 (initially present) and NO2 (initially formed) appears light brown. • When enough NO2 is formed, it can react to form N2O4: • 2NO2(g)  N2O4(g).

6. At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4: • The double arrow implies the process is dynamic. • Consider • Forward reaction: A  B Rate = kf[A] • Reverse reaction: B  A Rate = kr[B] • At equilibrium kf[A] = kr[B].

7. For an equilibrium we write • As the reaction progresses • [A] decreases to a constant, • [B] increases from zero to a constant. • When [A] and [B] are constant, equilibrium is achieved. • Alternatively: • kf[A] decreases to a constant, • kr[B] increases from zero to a constant. • When kf[A] = kr[B] equilibrium is achieved.

8. 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.

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

11. 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

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

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

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

15. [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

16. The Equilibrium Constant

17. 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

18. Law of mass action • Equilibrium constant expression • It depends on the reaction stoichiometry and NOT on its mechanism. • We can not tell the reaction mechanism from the Equilibrium Expression • Always products on the numerator • Always reactants in the denominator

19. What Are the Equilibrium Expressions for These Equilibria?

20. (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

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

22. 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)

23. 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.

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

25. 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.

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

27. 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.

28. January 11 • Manipulating Equilibrium Constants • Heterogeneous equilibrium • Equilibrium calculations RICE!!! • HW 23, 27 to 41 RED

29. 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.

30. 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.

31. 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.

32. Heterogeneous Equilibrium

33. 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.

34. 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

35. 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.

36. Writing Equilibrium Constant Expressions • The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. • The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. • The equilibrium constant is a dimensionless quantity. • In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. • If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

37. Calculating Equilibrium Concentrations • Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. • Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. • Having solved for x, calculate the equilibrium concentrations of all species. 14.4

38. Equilibrium Calculations

39. H2 (g) + Cl2 (g) 2 HCl (g) Equilibrium Calculations A closed system initially containing 1.000 x 10−3 M H2 and 2.000 x 10−3 M Cl2 At 448C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HCl is 1.87 x 10−3 M. Calculate Kc at 448C for the reaction taking place, which is

40. What Do We Know?

41. [HCl] Increases by 1.87 x 10-3M

42. Stoichiometry tells us [H2] and [I2]decrease by half as much

43. We can now calculate the equilibrium concentrations of all three compounds…

44. Kc= [HCl]2 [H2] [Cl2] (1.87 x 10-3)2 (6.5 x 10-5)(1.065 x 10-3) = = 51 …and, therefore, the equilibrium constant

45. Example 2 • The kp for the formation of ammonia is 1.45 x 10 -5 at 500 0C . In an equilibrium mixture of the 3 gases at 500 0C the partial pressure of H2 is .928 atm and for N2 is .432 atm. Find the partial pressure for NH3 at equilibrium. • Find Kc from the value of Kp

46. 1.45 x 10 -5

47. January 14 • Predicting the direction of a reaction • Le Chatelier’s Principle • HW Page 660 Q 1, 4, 5, 6, 7, • Le Chatelier’s Principle 51, 53, 55

48. Reaction of the day • Sodium chloride solution reacts with silver nitrate

49. Demo • Add NH3 to test tube and observe • A complex ion forms

50. COMPLEX IONS • Transition metal ion with attached ligands • Ligands are lewis bases with lone pairs of electrons. • A lewis base is a substance that has an unshared pair of electrons in its structure • H2O (aqua) • NH3 (ammine) • Cl- (chloro) • CN- (cyano) • OH- (hydroxo)