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Investigating Chemical Reactions: Equilibrium and Rates

Understand the dynamics of chemical reactions with a focus on equilibrium constants, reaction rates, and the law of mass balance. Explore how to calculate equilibrium constants for various reactions and their relationship to concentrations. Learn about heterogeneous equilibria and the concept of activity in reactions.

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Investigating Chemical Reactions: Equilibrium and Rates

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  1. Investigating Chemical Reactions • N2O4 (g) ⇄2 NO2 (g) • Colorless brown

  2. Closed Container: Reversibility

  3. Groups of Molecules A state of Dynamic Equilibrium

  4. Initial: all NO2 Equilibrium Concentrations 2 NO2 N2O4 • 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)”

  5. Rates of Reaction • For simple, one step reaction, reaction rate is due to inherent reactivity and collision rate • Forward rate = kf [N2O4] • Reverse rate = kr [NO2]2

  6. Initial: all NO2 Rates change over Course of Reaction 2 NO2 N2O4

  7. Same Principle for all Reactions CO + H2O CO2 + H2

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

  9. Law of Mass Balance • Empirical law—with justification from kinetics • Example: N2O4 2 NO2 This derivation is a simplification, but the outcome generally applies Forward rate = reverse rate [N2O4] = = = Equilibrium Constant

  10. N2O4 (g) ⇄ 2 NO2 (g) Keq = [NO2]2/[N2O4] • If equilibrium [R] and [P] are known, Keq can be calculated. • Example: at equilibrium, • [NO2] = 1.50 x 10-2 M, • [N2O4] = 1.03 x 10-2 M @ 317K (from experiment) Keq = (1.50 x 10-2)2/(1.03 x 10-2) = 0.0218

  11. Keq depends on how the equation is written. Small Keq (less than 1) means less product at equilibrium • N2O4 (g) ⇄ 2 NO2 (g) • K1 = [NO2]2/[N2O4] = 0.0218 • 2 NO2 (g) ⇄ N2O4 (g) • K2 = [N2O4] / [NO2]2 • K2 = 1/K1 • K2 = 1 / 0.0218 = 45.9 @ 317K • ½ N2O4 (g) ⇄ NO2 (g) • K3 = [NO2]/[N2O4]1/2 • K3 = (K1)1/2 • K3 = (0.0218)1/2 = 0.148 Large Keq (more than 1) means more product at equilibrium

  12. Test Your Understanding CO + H2O CO2 + H2 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 CO2 and H2. Answer: Keq = 0.022 Units?

  13. Keq 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 • Keq describes ratio, not absolute concentrations Concentrations are dependent on starting point, Ratio of concentrations is not!

  14. Initial: all NO2 all N2O4 both One Equilibrium Constant, Many Equilibrium Positions All proceed until equilibrium is reached

  15. Multiple Equilibria What if the products of one reaction act as reactants in a subsequent reaction? 2 NO2 ⇄ N2O4 K1 = [N2O4] / [NO2]2 N2O4 + O2⇄ 2 NO3 K2 = [NO3]2 / [N2O4][O2] 2 NO2 + O2⇄ 2 NO3 K3 = [NO3]2 / [NO3]2 [O2] The overall Keq can be calculated from individual steps. K3 = K1K2

  16. Units of Concentration • N2O4 (g) ⇄ 2 NO2 (g) • Concentrations may be measured in molarity or pressure • Kc = [NO2]2/[N2O4] --measured in molarity • Kp = PNO22 / PN2O4--measured in pressure

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

  18. Converting Between Kc and Kp • For N2O4 (g) ⇄ 2 NO2 (g) @ 317 K • Keq = (1.50 x 10-2)2/(1.03 x 10-2) = 0.0218 • Δn = 1 • Kp = 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: NO2 + CO  NO + CO2)

  19. 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

  20. 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) • Kc= [Ni(CO)4] / [CO]4

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

  22. 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

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

  24. Q Keq 2.0 45.9 2 NO2 (g) ⇄ N2O4 (g) Keq = 45.9 @ 317K [NO2] = 0.50 M [N2O4] = 0.50 M • Q = [N2O4] / [NO2]2 = (0.50)/ (0.50)2 = 2.0 Q = Keq , at equilibrium Q < Keq , moves to product (as written) Q > Keq , moves to reactant (as written)

  25. Equilibrium Calculations • To find Keq, equilibrium concentrations measured. • Practically, only one [reactant] or [product] measured • To calculate Keq, 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.

  26. 1.00 mol of CO and 1.00 mol of H2O are placed in a 50.0 L vessel. At equilibrium, [CO2] is found to be 0.0086 M @ 1273 K. Calc. Keq. CO (g) + H2O (g) ⇄ CO2 (g) + H2 (g) I C E 1.00/50.0 = 0.0200 0.0200 0 0 - 0.0086 - 0.0086 + 0.0086 + 0.0086 0.0200 - 0.0086 = 0.0114 0.0200 - 0.0086 = 0.0114 0.0086 0.0086 Keq = [CO2][H2] = (0.0086)(0.0086) = 0.569 [CO][H2O] (0.0114)(0.0114)

  27. What are the equilibrium concentrations of each species if 0.500 mol of H2 and I2 are placed into a 1.00 L vessel at 458 oC? H2 (g) + I2 (g) ⇄ 2 HI (g) Keq = 49.7 @ 458 oC

  28. PCl5 (g) ⇄ PCl3 (g) + Cl2 (g) If the initial concentration of PCl5 is 1.00 M, calculate the equilibrium concentration of each species at 160 oCif K = 0.0211 at that temperature.

  29. 0.030 mol of SO2Cl2, 2.0 mol of SO2 and 1.0 mol of Cl2 are placed into a 100 L reaction vessel at 173 oC. The K for that temperature is 0.082. Find the equilibrium concentrations of all species. SO2Cl2 (g) ⇄SO2 (g) + Cl2 (g)

  30. A 2.00 L container at 463 K contains 0.500 mol of phosgene, COCl2. K is 4.93 x 10-3 for COCl2 (g) ⇄CO (g) + Cl2 (g) Calculate the equilibrium concentrations of all species.

  31. Calculate the partial pressure of SO3, given Kp is 0.74 (at 2100 K) for CaSO4 (s) ⇄CaO (s) + SO3 (g)

  32. 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

  33. Change in Concentration • 2 NO2 (g) ⇄ N2O4 (g) • Add NO2 (either conc. or pressure) • Equilibrium shifts to right (more products) • Remove NO2 • Equilibrium shifts to left • Add N2O4 • Equilibrium shifts to left Logic: Think about changing rates

  34. Fig. 14.7 Example: Haber Process

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

  36. Applications of Le Chatelier • Strategies for driving unfavorable metabolic reactions

  37. LeChatelier’s Principle • Adding/removing gases not involved in the equilibrium – has no effect on system—no change in partial pressures of reactants • 2 NO2 (g) ⇄ N2O4 (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

  38. Changes in Temperature • Changing temperature changes equilibrium position and the equilibrium constant (Keq) • CO2 (g) + C (s) ⇄ 2 CO (g) ΔH = +173 kJ • Endothermic--re-write equation as: • HEAT + CO2 (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

  39. Test Your Understanding N2O4 (colorless) 2 NO2 (brown) Is this reaction exothermic or endothermic as written?

  40. Haber Process • Production of ammonia is big business! • Given the equation and data, how would you run the process to maximize ammonia output? N2 (g) + 3H2 (g)  2NH3 (g)

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

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

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

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

  45. Test Your Understanding • Write an aqueous acid/base reaction for CH4. Then write the equilibrium expression. (The expression for Ka.)

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