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Chapter 14

Chapter 14. Chemical Equilibrium. Chemical Equilibrium. condition of a reversible reaction where the rate of the forward reaction is equal to the rate of the reverse reaction a dynamic condition, meaning that the reaction continues with no net change in the amounts of reactants and products.

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Chapter 14

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  1. Chapter 14 Chemical Equilibrium

  2. Chemical Equilibrium • condition of a reversible reaction where the rate of the forward reaction is equal to the rate of the reverse reaction • a dynamic condition, meaning that the reaction continues with no net change in the amounts of reactants and products

  3. Chemical Equilibrium • products predominate over reactants • product-favored • reactants predominate over products • reactant-favored

  4. Reactants, Products, and Equilibrium

  5. Equilibrium • independent of direction of approach • catalyst only changes how fast equilibrium is reached

  6. Approach to Equilibrium

  7. Law of Mass Action aA + bB + cC + ... pP + qQ + rR + ... Equilibrium Constant [P]p [Q]q [R]r ... Kc = [A]a [B]b [C]c ...

  8. Values of Kc • value of equilibrium constant of a reaction, and that for its reverse reaction,are reciprocals of each other

  9. Gas Phase Equilibrium 2 NO2(g)  N2O4(g) [N2O4] Kc = [NO2]2

  10. Heterogeneous Equilibrium CaCO3(s) + heat  CaO(s) + CO2(g) [CaO(s)][CO2(g)] Kc = [CaCO3(s)] - concentrations of pure solids and liquids are constant are dropped from expression Kc = [CO2(g)]

  11. Acid Dissociation Constant HC2H3O2 + H2O  H3O+ + C2H3O2- [H3O+][C2H3O2-] K = [H2O][HC2H3O2] [H3O+][C2H3O2-] Ka = K  [H2O] = [HC2H3O2]

  12. Base Dissociation Constant NH3 + H2O  NH4+ + OH- [NH4+][OH-] K = [H2O][NH3] [NH4+][OH-] Kb = K  [H2O] = [NH3]

  13. Autoionization of Water H2O + H2O  H3O+ + OH- [H3O+][OH-] K = [H2O]2 Kw = K [H2O]2 = [H3O+][OH-] = 1.0  10-14

  14. Stepwise Equilibrium [NO]2 Kc1 = [N2][O2] (1.) N2(g) + O2(g) 2NO(g) [NO2]2 Kc2 = [NO]2[O2] (2.) 2NO(g) + O2(g) 2NO2(g) Combine (1.) & (2.) N2(g) + 2O2(g) 2NO2(g) [NO]2 [NO2]2 Kc =  = Kc1 Kc2 [N2][O2] [NO]2[O2]

  15. Ideal Gas Law PV = nRT P = (n/V)RT PA = [A]RT [A] = PA/RT

  16. Pressure Equilibrium Constants [NH3]2 Kc = [N2][H2]3 N2 + 3H2 2NH3 (PNH3/RT)2 = (PN2/RT) (PH2/RT)3 PNH32 (1/RT)2 Kc = (PN2(1/RT))(PH23(1/RT)3) PNH32 (1/RT)2 = PN2 PH23(1/RT)(1/RT)3 (1/RT)2 = Kp (1/RT)(1/RT)3

  17. Kc vs. Kp N2 + 3H2 2NH3 (1/RT)2 Kc = Kp = Kp (1/RT)-2 (1/RT)(1/RT)3 • In General • Kc = Kp (1/RT)Dn • where Dn = #moles gaseous products • - # moles gaseous reactants

  18. Kp vs. Kc In General Kp = Kc (RT)Dn where Dn = #moles gaseous products - # moles gaseous reactants

  19. Determining Equilibrium Constants 1. Derive the equilibrium constant expression for the balanced chemical equation 2. Construct a Reaction Table with information about reactants and products 3. Include the amounts reacted, x, in the Reaction Table 4. Calculate the equilibrium constant in terms of x

  20. Determining Equilibrium Constants 5. Solve equation for x 6. Check for reasonableness

  21. N2O4(g) 2 NO2(g)

  22. Example: An equilibrium is established by placing 2.00 moles of N2O4(g) in a 5.00 L and heating the flask to 407 K. It was determined that at equilibrium the concentration of the NO2(g) is 0.525 mol/L. What is the value of the equilibrium constant? N2O4(g) 2 NO2(g) [NO2]2 Kc = [N2O4] N2O4(g) 2 NO2(g) [Initial] {mol/L) 0.40 0 [change] [Equilibrium] 0.525 x -1/2x 0.40 - 1/2x = 0 + x

  23. Example: An equilibrium is established by placing 2.00 moles of N2O4(g) in a 5.00 L and heating the flask to 407 K. It was determined that at equilibrium the concentration of the NO2(g) is 0.525 mol/L. What is the value of the equilibrium constant? 0.525 = 0 + x [NO2]2 Kc = [N2O4] 0.40 - 1/2x x = 0.525 [NO2] = 0.40 - 1/2x = 0.40 - 1/2(0525) [NO2]2 (0.525)2 Kc = = [N2O4] 0.138 = 0.138 = 2.00

  24. Meaning of Equilibrium Constant • K>>1: reaction is product-favored; equilibrium concentrations of products are greater than equilibrium concentrations of reactants. • K<<1: reaction is reactant-favored; equilibrium concentrations of reactants are greater than equilibrium concentrations of products.

  25. Selected Equilibrium Constants

  26. Concentrations ofReactants and Products

  27. Predicting the Direction of a Reaction

  28. Example: An equilibrium is established by placing 2.00 moles of N2O4(g) in a 5.00 L at 298 K. What are the equilibrium concentrations ofN2O4(g) and NO2(g)? Kc = 1.70 x 102 at 298 K. N2O4(g) 2 NO2(g) [NO2]2 Kc = = 170 [N2O4] N2O4(g) 2 NO2(g) [Initial] {mol/L) 0.400 0 [change] [Equilibrium] 2x -x 0.400 - x 2x

  29. Example: An equilibrium is established by placing 2.00 moles of N2O4(g) in a 5.00 L at 298 K. What are the equilibrium concentrations ofN2O4(g) and NO2(g)? Kc = 1.70 x 102 at 298 K. [NO2]2 Kc = = 170 [N2O4] (2x)2 = (0.40 - x) x = 0.396 mol/L 68 - 170x = 4x2 [NO2] = 2(0.396 mol/L) = 0.792 mol/L [N2O4] = 0.400 - 0.396 = 0.004 mol/L

  30. Le Chatelier's Principle If a stress, such as a change in concentration, pressure, temperature, etc., is applied to a system at equilibrium, the equilibrium will shift in such a way as to lessen the effect of the stress.

  31. Production of Ammonia catalysis N2(g) + 3 H2(g) 2 NH3(g) + heat high pressure and temperature

  32. Haber-Bosch Process

  33. Increase in Concentrationor Partial Pressure for N2(g) + 3 H2(g) 2 NH3(g) an increase in N2 and/or H2 concentration or pressure, will cause the equilibrium to shift towards the production of NH3

  34. Decrease in Concentration or Partial Pressure for N2(g) + 3 H2(g) 2 NH3(g) likewise, a decrease in NH3 concentration or pressure will cause more NH3 to be produced

  35. Changes in Temperature for N2(g) + 3 H2(g) 2 NH3(g) + heat for an exothermic reaction, an increase in temperature will cause the reaction to shift back towards reactants

  36. Increase in Volume for N2(g) + 3 H2(g) 2 NH3(g) + heat an increase in volume, causes the equilibrium to shift to the left where there are more gaseous molecules

  37. Decrease in Volume for N2(g) + 3 H2(g) 2 NH3(g) + heat a decrease in volume, causes the equilibrium to shift to the right where there are fewer gaseous molecules

  38. Stress on Equilibrium

  39. Shifting of Equilibrium

  40. Probability andChemical Equilibrium

  41. Entropy • measure of the disorder in the system • more disorder for gaseous systems than liquid systems, more than solid systems

  42. Equilibrium Systems • product-favored if K > 1 • exothermic reactions favor products • increasing entropy in system favors products • at low temperature, product-favored reactions are usually exothermic • at high temperatures, product-favored reactions usually have increase in entropy

  43. Reaction Rates • reactions occur faster in gaseous phase than solids and liquids • reactions rates increase as temperature increases • reactions rates increase as concentration increases • rates increase as particle size decreases • rates increase with a catalyst

  44. Ammonia Synthesis • reaction is slow at room temperature, raising temperature, increases rate but lowers yield • increasing pressure shifts equilibrium to products • liquefying ammonia shifts equilibrium to products • use of catalyst increases rate

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