2. Investigating Chemical Reactions N 2 O 4 (g) ⇄ 2 NO 2 (g) Colorless brown

3. Closed Container: Reversibility

4. Groups of Molecules A state of Dynamic Equilibrium

5. Equilibrium Concentrations Equilibrium is reached when the concentrations of products and reactants remains unchanged with time Condition dependent Qualitative descriptions –“Equilibrium lies to the left (or right)” –“Equilibrium favors products (or reactants)” Initial: all NO 2 2 NO 2 N 2 O 4

6. Rates of Reaction For simple, one step reaction, reaction rate is due to inherent reactivity and collision rate Forward rate = k f [N 2 O 4 ] Reverse rate = k r [NO 2 ] 2

7. Rates change over Course of Reaction Initial: all NO 2 2 NO 2 N 2 O 4

8. Same Principle for all Reactions CO + H 2 O CO 2 + H 2

9. Law of Mass Balance General, Empirical Form: aA + bB ⇄ cC + dD This is called the Equilibrium expression

10. Law of Mass Balance Empirical law—with justification from kinetics Example: N 2 O 4 2 NO 2 Forward rate = reverse rate This derivation is a simplification, but the outcome generally applies

11. N 2 O 4 (g) ⇄ 2 NO 2 (g) K eq = [NO 2 ] 2 /[N 2 O 4 ] If equilibrium [R] and [P] are known, K eq can be calculated. Example: at equilibrium, –[NO 2 ] = 1.50 x 10 -2 M, –[N 2 O 4 ] = 1.03 x 10 -2 M @ 317K (from experiment) K eq = (1.50 x 10 -2 ) 2 /(1.03 x 10 -2 ) = 0.0218

12. K eq depends on how the equation is written. N 2 O 4 (g) ⇄ 2 NO 2 (g) –K 1 = [NO 2 ] 2 /[N 2 O 4 ] = 0.0218 2 NO 2 (g) ⇄ N 2 O 4 (g) –K 2 = [N 2 O 4 ] / [NO 2 ] 2 –K 2 = 1/K 1 –K 2 = 1 / 0.0218 = 45.9 @ 317K ½ N 2 O 4 (g) ⇄ NO 2 (g) –K 3 = [NO 2 ] /[N 2 O 4 ] 1/2 –K 3 = (K 1 ) 1/2 –K 3 = (0.0218) 1/2 = 0.148 Small Keq (less than 1) means less product at equilibrium Large Keq (more than 1) means more product at equilibrium

13. Test Your Understanding Calculate the equilibrium constant for at a given temperature if a 1-L container that initially held 1 mol of CO and 1 mol of water reached equilibrium and then had 0.13 mol of CO 2 and H 2. CO + H 2 O CO 2 + H 2 Answer: K eq = 0.022 Units?

14. K eq describes ratio of reactants and products A ⇄ B Keq =0.33 Which of the following systems are at equilibrium? –[A] = 3.0 M, [B] = 1.0 M –[A] = 7.5 x 10 -3 M, [B] = 2.5 x 10 -3 M –[A] = 12.0 M, [B] = 4.0 M K eq describes ratio, not absolute concentrations Concentrations are dependent on starting point, Ratio of concentrations is not!

16. Initial: all NO 2 All proceed until equilibrium is reached One Equilibrium Constant, Many Equilibrium Positions all N 2 O 4 both

17. 2 NO 2 ⇄ N 2 O 4 K 1 = [N 2 O 4 ] / [NO 2 ] 2 N 2 O 4 + O 2 ⇄ 2 NO 3 K 2 = [NO 3 ] 2 / [N 2 O 4 ][O 2 ] 2 NO 2 + O 2 ⇄ 2 NO 3 K 3 = [NO 3 ] 2 / [NO 3 ] 2 [O 2 ] K 3 = K 1 K 2 Multiple Equilibria What if the products of one reaction act as reactants in a subsequent reaction? The overall K eq can be calculated from individual steps.

18. N 2 O 4 (g) ⇄ 2 NO 2 (g) –Concentrations may be measured in molarity or pressure K c = [NO 2 ] 2 /[N 2 O 4 ] --measured in molarity K p = P NO 2 2 / P N 2 O 4 --measured in pressure Units of Concentration

19. Relationship between K p and K c –Describes same phenomenon in different ways, so quantity may be different –Example: A B –P =(n/V)RT = [C]RT –K p = K c (RT) Δn –where Δn = # mol gas product - # mol gas reactant

20. Converting Between K c and K p For N 2 O 4 (g) ⇄ 2 NO 2 (g) @ 317 K –K eq = (1.50 x 10 -2 ) 2 /(1.03 x 10 -2 ) = 0.0218 –Δn = 1 –K p = 0.0218[(0.08206)(317)] 1 = 0.567 What is the relationship between Kc and Kp when there is no change in number of moles of gas? (Ex: NO 2 + CO  NO + CO 2 )

21. The Concept of Activity More accurately, the equilibrium expression does not depend on the concentration, but the activity Activity = Concentration/Reference –Activity of 0.5 M SO2 = 0.5M/1 M = 0.5 –Activity of 2.7 atm CO2 = 2.7atm/1 atm = 2.7 –Activity of pure liquids and solids: reference is pure compound Activity of liquid water = 55M water/55M water = 1 Activities of all pure liquids and solids are unity

22. Heterogeneous Equilibria We have been talking about homogeneous equilibria – where all reactants and products are in the same state. (gases or solutions) What about mixed phases? Ni (s) + 4 CO (g) ⇄ Ni(CO) 4 (g) K c = [Ni(CO) 4 ] / [CO] 4

24. Summary of Equilibrium Expressions K eq tells us about the ratio of reactants and products at equilibrium (not absolute values) Must specify equation and temperature with K eq If equilibrium concentrations of all reactants and products are known, K eq may be determined. The K eq of multiple equilibria may be determined from the K eq s of individual reactions K c and K p are interconvertable Solids and pure liquids don’t affect K eq

25. Quantitative Equilibrium Problems Determine whether or not a reaction is at equilibrium Calculate an equilibrium constant from equilibrium concentrations Given starting concentrations and K, predict the equilibrium positions

26. Reaction Quotient Reaction quotient – Q – the same expression as K eq but with the current concentrations (not equilibrium concentrations) aA + bB ⇄ cC + dD Q = [C] c [D] d [A] a [B] b At equilibrium, Q = K eq Q helps to determine the direction of the reaction Reactions move toward equilibrium Q  K eq

27. 2 NO 2 (g) ⇄ N 2 O 4 (g) K eq = 45.9 @ 317K [NO 2 ] = 0.50 M [N 2 O 4 ] = 0.50 M Q = [N 2 O 4 ] / [NO 2 ] 2 = (0.50)/ (0.50) 2 = 2.0 Q = K eq, at equilibrium Q < K eq, moves to product (as written) Q > K eq, moves to reactant (as written) K eq 45.9 Q 2.0

28. Equilibrium Calculations To find K eq, equilibrium concentrations measured. –Practically, only one [reactant] or [product] measured To calculate K eq, we will need the equilibrium expression and 3 other pieces of information. –Initial Concentrations –Changes due to reaction – depends on stoichiometry –Equilibrium concentrations To do any equilibrium calculation, we will need to set up an ICE table.

29. 1.00 mol of CO and 1.00 mol of H 2 O are placed in a 50.0 L vessel. At equilibrium, [CO 2 ] is found to be 0.0086 M @ 1273 K. Calc. K eq. CO (g) + H 2 O (g) ⇄ CO 2 (g) + H 2 (g) ICEICE 1.00/50.0 = 0.0200 0.020000 0.0086 + 0.0086 0.0086 - 0.0086 0.0200 - 0.0086 = 0.0114 K eq = [CO 2 ][H 2 ] = (0.0086)(0.0086) = 0.569 [CO][H 2 O] (0.0114)(0.0114)

30. What are the equilibrium concentrations of each species if 0.500 mol of H 2 and I 2 are placed into a 1.00 L vessel at 458 o C? H 2 (g) + I 2 (g) ⇄ 2 HI (g) K eq = 49.7 @ 458 o C

31. PCl 5 (g) ⇄ PCl 3 (g) + Cl 2 (g) If the initial concentration of PCl 5 is 1.00 M, calculate the equilibrium concentration of each species at 160 o C if K = 0.0211 at that temperature.

32. 0.030 mol of SO 2 Cl 2, 2.0 mol of SO 2 and 1.0 mol of Cl 2 are placed into a 100 L reaction vessel at 173 o C. The K for that temperature is 0.082. Find the equilibrium concentrations of all species. SO 2 Cl 2 (g) ⇄ SO 2 (g) + Cl 2 (g)

33. A 2.00 L container at 463 K contains 0.500 mol of phosgene, COCl 2. K is 4.93 x 10 -3 for COCl 2 (g) ⇄ CO (g) + Cl 2 (g) Calculate the equilibrium concentrations of all species.

34. Calculate the partial pressure of SO 3, given K p is 0.74 (at 2100 K) for CaSO 4 (s) ⇄ CaO (s) + SO 3 (g)

35. LeChatelier’s Principle “If a stress is applied to a system at equilibrium, the system reacts to relieve the stress.” Sets up new equilibrium position Change in –Concentration of reactant or product –Pressure of system –Temperature of system In this case, changes equilibrium position AND equilibrium constant

36. Change in Concentration 2 NO 2 (g) ⇄ N 2 O 4 (g) Add NO 2 (either conc. or pressure) –Equilibrium shifts to right (more products) Remove NO 2 –Equilibrium shifts to left Add N 2 O 4 –Equilibrium shifts to left Logic: Think about changing rates

38. Fig. 14.7 Example: Haber Process

39. Applications of Le Chatelier “Forcing” an unfavorable reaction

40. Applications of Le Chatelier Strategies for driving unfavorable metabolic reactions

41. LeChatelier’s Principle Adding/removing gases not involved in the equilibrium – has no effect on system—no change in partial pressures of reactants 2 NO 2 (g) ⇄ N 2 O 4 (g) Decrease the volume? –A decrease in volume will favor the side with the least # moles of gas –An increase in volume will favor the side with the greater # moles of gas –If moles of gas equal, no effect

43. Changes in Temperature Changing temperature changes equilibrium position and the equilibrium constant (K eq ) CO 2 (g) + C (s) ⇄ 2 CO (g) ΔH = +173 kJ –Endothermic--re-write equation as: HEAT + CO 2 (g) + C (s) ⇄ 2 CO (g) Follow Le Chatelier –Add heat (inc temp), increases K, shifts to right (products) –Remove heat (dec temp), decreases K, shift to left

44. N 2 O 4 (colorless) 2 NO 2 (brown) Test Your Understanding Is this reaction exothermic or endothermic as written?

45. Haber Process Production of ammonia is big business! Given the equation and data, how would you run the process to maximize ammonia output? N 2 (g) + 3H 2 (g)  2NH 3 (g)

47. Equilibria of Processes Equilibrium can describe process as well as reaction Conformation Solubility

48. Acid-Base Chemistry Major application of equilibrium (ch 7-8) Acid/base reactions reach equilibrium quickly Relatively simple reaction with MAJOR applications

49. Acid/Base Reactions Bronsted-Lowry Definition –Acids are proton (H+) donors –Bases are proton (H+) acceptors (lone pair) Limit discussion to aqueous solutions

50. Acid-Base Reactions Base Acid Conjugate acid Conj. Base

51. Test Your Understanding Write an aqueous acid/base reaction for CH 4. Then write the equilibrium expression. (The expression for K a.)

52. Defining Strong and Weak Acids K a of CH 4 = 10 -50 K a of HC 2 H 3 O 2 = 10 -5 K a of HCl = 10 7 In which direction does each equilibrium lie? What is a strong acid? What is a weak acid? What is a “stronger” and “weaker” acid?

53. Defining Strong and Weak Bases

54. Strong Acids H 2 SO 4 HNO 3 HClO 4 HCl, HBr, HI At equilibrium, [H + ] = [HA] o

55. Test Your Concept

56. Working with pKa pKa is convenient way to express Ka pKa = -log Ka KapKa 1 x 10 -12 1 x 10 8 35 -9 4.76 x 10 -5 How does pKa value relate to acid strength?

57. Weak Acid-Base Equilibrium: Nothing New What is the % dissociation of the side chain of aspartate in a 0.600M aq. solution (pKa 3.9)? Answer: 1.4% ionized

58. pH of Aqueous Solutions Neutral water: autoionization Kw = 1 x 10 -14 [HO - ] = [H + ] = 10 -7 pH = 7 In concentrated acid solutions, we assume autoionization is a minor contributor: “all” hydronium from acid dissociation

59. What is the pH of a 0.45M aq solution of acetic acid (pKa = 5.0)? 1. Write equilibrium expression 2. Solve ICE to determine [H+] 3. Determine pH Answer: 2.7

60. What happens to the pH of a 0.45M aq solution of acetic acid (pKa = 5.0) if sodium acetate is added to give an acetate concentration of 0.05M? What happens to the pH of a 0.45M aq solution of acetic acid (pKa = 5.0) if sodium acetate is added to give an acetate concentration of 0.45M? What is the pH of a 0.50 M aq solution of acetic acid (pKa = 5.0) to which has been added 0.05M NaOH? Answers: 4.0, 5.0, 4.0