1 / 52

Equilibrium

Equilibrium. Equilibrium. Limiting reagent Concentrations become constant Dynamic situation Reversible reactions. 2NO 2  N 2 O 4. a. b. c. d. time. 2NO 2  N 2 O 4. Equilibrium Constant (K). represents a ratio of the concentrations of products to reactants at equilibrium:

zasha
Download Presentation

Equilibrium

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Equilibrium

  2. Equilibrium • Limiting reagent • Concentrations become constant • Dynamic situation • Reversible reactions

  3. 2NO2 N2O4 a b c d time

  4. 2NO2 N2O4

  5. Equilibrium Constant (K) represents a ratio of the concentrations of products to reactants at equilibrium: aA +bB  cC + dD

  6. Equilibrium, or K, expression Equilibrium Constant (K) [C]c[D]d [A]a[B]b K=

  7. Equilibrium Constant (K) • “[]” represents concentration in mol/L for (g) and (aq), only • Each “[]” must be raised to the power of its coefficient

  8. Equilibrium Constant (K) • K < 1 indicates little product formation • K > 100 indicates great amount of product formation

  9. Equilibrium Constant (K) • Write the K expression for the dimerization of nitrogen dioxide.

  10. Equilibrium Constant (K) [N2O4] [NO2]2 What will the units of K be in this example? L/mol K=

  11. Equilibrium Constant (K) • At 25°C, the equilibrium concentrations of NO2 and N2O4 are 0.0370M and 0.2315M. What is the value of K at this T?

  12. Equilibrium Constant (K) [0.2315] [0.037]2 K= K=0.2315mol/L 0.001369mol2/L2 K=169 L/mol

  13. N2 + 3H2 2NH3 • Write the K expression for the synthesis of ammonia.

  14. Equilibrium Constant (K) [NH3]2 [N2][H2]3 What will the units of K be in this example? L2/mol2 K=

  15. Equilibrium Constant (K) • At 300°C, the equilibrium concentrations are: [N2]eq= 2.59M [H2]eq=2.77M [NH3]eq=1.82M What is the value of K at this temperature?

  16. Equilibrium Constant (K) [1.82]2 [2.59][2.77]3 K= K=3.3124mol2/L2 55.05mol4/L4 K=0.0602 L2/mol2

  17. Equilibrium Constant (K) • Small K (<1) means… • Big K (>100) means… • Different manner of solving problems

  18. Equilibrium Constant (K) • If a reaction is reversed, then the value of K for the reversed reaction is the reciprocal of K.

  19. Equilibrium Constant (K) • So, if the dimerization of NO2 is reversed to be the decomposition of N2O4…

  20. Equilibrium Constant (K) • K = (169 L/mol)-1 or 0.00592 mol/L

  21. Equilibrium Constant (K) • At 25°C, the initial concentration of N2O4 is 0.750M. What are the eq. conc. of both species at this temperature?

  22. Equilibrium Constant (K) • You will make an equilibrium chart to indicate the initial, change, and equilibrium concentrations.

  23. Equilibrium Constant (K)

  24. Equilibrium Constant (K)

  25. Equilibrium Constant (K)

  26. [2x]2 [0.75] 0.00592= 4.44 x 10-3= 4x2 Equilibrium Constant (K) [2x]2 [0.75 – x] 0.00592=

  27. 0.0333= x Equilibrium Constant (K) 1.11 x 10-3= x2

  28. Equilibrium Constant (K) 5% rule…is what you removed less than 5% of the smaller initial value?

  29. Equilibrium Constant (K) If so, then your assumption that what you removed was so small it is negligible is correct

  30. Equilibrium Constant (K) 5% rule test: 1x 1(0.0333) X 100 = 0.75 4.44% < 5%… assumption is good

  31. Equilibrium Constant (K)

  32. Equilibrium Constant (K) Check your answer: [0.0666]2 [0.7167] K= 0.00619 K=

  33. Equilibrium Constant (K) • At 25°C, the initial concentration of NO2 is 0.500M. What are the eq. conc. Of both species at this temperature? Remember that K = 169L/mol.

  34. Equilibrium Constant (K) Since K is big, lots of product will be made. Thus, almost all of the initial amount of reactant will be used. You need to make two charts for a big K problem…Stoichiometry and Equilibrium

  35. Equilibrium Constant (K)Stoichiometry Chart

  36. Equilibrium Constant (K) Stoichiometry Chart

  37. Equilibrium Constant (K) Stoichiometry Chart

  38. Equilibrium Constant (K) Equilibrium Chart

  39. Equilibrium Constant (K) Equilibrium Chart

  40. Equilibrium Constant (K) Equilibrium Chart

  41. [0.25] 4x2 169= 676x2= 0.25 Equilibrium Constant (K) [0.25 - x] [2 x]2 169=

  42. 0.0192= x Equilibrium Constant (K) 3.70 x 10-4= x2

  43. Equilibrium Constant (K) 5% rule test: 1x 1(0.0192) X 100 = 0.25 7.69% > 5%… assumption is bad

  44. Equilibrium Constant (K) Since the assumption that x was so small it was negligible is bad, then you must re-insert x and solve the equation with x present.

  45. [0.25 - x] 4x2 169= Equilibrium Constant (K) [0.25 - x] [2 x]2 169= 676x2= 0.25 - x

  46. x = -b ± b2 – 4ac 2a Equilibrium Constant (K) 676x2 + x – 0.25 = 0 ax2 + bx + c = 0

  47. Equilibrium Constant (K) You will get two values of x. If both are positive, then you will always select the smaller one. If one is positive and the other negative, you will select the positive one.

  48. x = -1 ± 12 – 4(676)(-0.25) 2(676) x = -1 ± 1 – (-676) 1352 Equilibrium Constant (K)

  49. x = -1 ± 677 1352 Equilibrium Constant (K) x = -1 ± 26.02 1352

  50. Equilibrium Constant (K) x = 25.02 1352 = 0.0185 *accept this one OR x = -27.02 1352 = -0.0200

More Related