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Drill:

Drill:. List five factors & explain how each affect reaction rates. Drill: Solve Rate Law. A + B C + D fast 4 C + A 2G fast 2 K 4D + B fast G + K 2 Q + 2 W fast Q + W Prod. slow. Review Drill & Check HW. CHM II HW. Review PP-19 & 20

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Drill:

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  1. Drill: • List five factors & explain how each affect reaction rates

  2. Drill: Solve Rate Law A + B C + D fast 4 C + A 2G fast 2 K 4D + B fast G + K 2 Q + 2 W fast Q + W Prod. slow

  3. Review Drill&Check HW

  4. CHM II HW • Review PP-19 & 20 • Complete the attached worksheet & turn it in tomorrow • Lab Thursday

  5. Chemical Equilibria

  6. Equilibrium • The point at which the rate of a forward reaction = the rate of its reverse reaction

  7. Equilibrium • The concentration of all reactants & products become constant at equilibrium

  8. Equilibrium • Because concentrations become constant, equilibrium is sometimes called steady state

  9. Equilibrium • Reactions do not stop at equilibrium, forward & reverse reaction rates become equal

  10. Reaction • aA(aq)+ bB(aq) pP(aq)+ qQ(aq) • Ratef = kf[A]a[B]b • Rater = kr[P]p[Q]q • At equilibrium, Ratef = Rater • kf[A]a[B]b = kr[P]p[Q]q

  11. At equilibrium, Ratef = Rater kf[A]a[B]b = kr[C]c[D]d kf /kr = ([C]c[D]d)/ ( [A]a[B]b) kf /kr = Kc = Keq in terms of concentration Kc = ([C]c[D]d)/ ( [A]a[B]b)

  12. Reaction aA(g)+ bB(g)<-->cC(g)+ dD(g) Ratef = kfPAaPBb Rater = krPCcPDd At equilibrium, Ratef = Rater kfPAaPBb = krPCcPDd

  13. At equilibrium, Ratef = Rater kfPAaPBb = krPCcPDd kf /kr = (PCcPDd)/ ( PAaPBb) kf /kr = Kp = Keq in terms of pressure Kp = (PCcPDd)/ ( PAaPBb)

  14. All Aqueous aA + bB pP+ qQ

  15. Equilibrium Expression ( Products)p (Reactants)r Keq=

  16. AP CHM HW • Read: Chapter 12 • Work problems: 5, 7, & 12 • Page: 365

  17. CHM II HW • Read: Chapter 17 • Work problems: 17 & 21 • Page: 745

  18. Equilibrium Applications • When K >1, [p] > [r] • When K <1, [p] < [r]

  19. Equilibrium Calculations • Kp = Kc(RT)Dngas

  20. Equilibrium Expression • Reactants or products not in the same phase are not included in the equilibrium expression

  21. Equilibrium Expression aA(s)+ bB(aq)<--> cC(aq)+ dD(aq) [C]c [D]d [B]b Keq=

  22. Reaction Mechanism • Sequence of steps that make up the total reaction process

  23. Reaction Mechanism • 1) A + B <---> C Fast • 2) A + C <---> D Fast • 3) B + D <---> H Fast • 4) H + A -----> P Slow

  24. Reaction Mechanism • The rate determining step is the slowest step • H + A ----> P Slow • Rate = k4[H][A]

  25. Reaction Mechanism • Rate = k4[H][A] • Because H is not one of the original reactants, H cannot be used in a rate expression

  26. Reaction Mechanism • 3) B + D <---> H • K3 = [H]/([B][D]) • [H] = K3[B][D]

  27. Reaction Mechanism • [H] = K3[B][D] • Rate = k4[H][A] • Rate = k4K3[B][D][A]

  28. Reaction Mechanism • 2) A + C <---> D • K2 = [D]/([A][C]) • [D] = K2[A][C]

  29. Reaction Mechanism • [D] = K2[A][C] • Rate = k4K3[B][D][A] • Rate = k4K3[B]K2[A][C][A] • Rate = k4K3 K2[B][A]2[C]

  30. Reaction Mechanism • 1) A + B <---> C • K1 = [C]/([A][B]) • [C] = K1[A][B]

  31. Reaction Mechanism • [C] = K1[A][B] • Rate = k4K3 K2[B][A]2[C] • Rate = k4K3 K2[B][A]2K1[A][B] • Rate = k4K3 K2K1 [B]2[A]3 • Rate = K[B]2[A]3

  32. Solve Rate Expression • 1) A + B <---> 2C Fast • 2) A + C <---> D Fast • 3) B + D <---> 2H Fast • 4) 2H + A ----> P Slow

  33. Reaction Mechanism • When one of the intermediates anywhere in a reaction mechanism is altered, all intermediates are affected

  34. Reaction Mechanism • 1) A + B <---> C + D • 2) C + D <---> E + K • 3) E + K <---> H + M • 4) H + M <----> P

  35. Lab Results • % 100 80 60 40 • RT 5.21 8.42 11.9 21.7 • WR 2.75 4.23 7.96 11.2

  36. Applications of Equilibrium Constants where [A], [B], [P], and [Q] are molarities at any time. Q = K only at equilibrium.

  37. NH3 H2 + N2 At a certain temperature at equilibrium Pammonia = 4.0 Atm, Phydrogen = 2.0 Atm, & Pnitrogen = 5.0 Atm. Calculate Keq:

  38. Equilibrium Applications • When K > Q, the reaction goes forward • When K < Q, the reaction goes in reverse

  39. Drill: SO2 + O2 SO3 • Determine the magnitude of the equilibrium constant if the partial pressure of each gas is 0.667 Atm.

  40. Review Drill&Collect HW

  41. CHM II HW • Review PP-20 • Complete the attached worksheet & turn it in Friday • Lab Tomorrow

  42. Equilibrium Calculations • aA + bB <--> cC + dD • Stoichiometry is used to calculate the theoretical yield in a one directional rxn

  43. Equilibrium Calculations • aA + bB <--> cC + dD • In equilibrium rxns, no reactant gets used up; so, calculations are different

  44. Equilibrium Calculations • Set & balance rxn • Assign amounts • Write eq expression • Substitute amounts • Solve for x

  45. Equilibrium Calculations • CO + H2O CO2 + H2 • Calculate the partial pressure of each portion at eq.when 100.0 kPa CO & 50.0 kPa H2O are combined: • Kp = 3.4 x 10-2

  46. Equilibrium Calculations • CO H2O CO2 H2 • 100 -x 50 - x x x • Kp =PCO2PH2 = x2 • PCOPH2O(100-x) (50-x) • Kp = 3.4 x 10-2

  47. Equilibrium Calculations x2 x2 (100 -x)(50 - x) = 5000 -150x + x2 = 3.4 x 10-2 x2 = 170 - 5.1x + 0.034x2 0.966x2 + 5.1x - 170 = 0

  48. Equilibrium Calculations Xe (g) + F2(g) XeF2(g) Calculate the partial pressure of each portion when 50.0 kPa Xe & 100.0 kPa F2 are combined: Kp = 4.0 x 10-4

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