1. Equilibrium

2. Limiting reagent Concentrations become constant Dynamic situation Reversible reactions

3. 2NO 2  N 2 O 4 abcd time

4. 2NO 2  N 2 O 4

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

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

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) [N 2 O 4 ] [NO 2 ] 2 What will the units of K be in this example? K= L/mol

11. Equilibrium Constant (K) At 25°C, the equilibrium concentrations of NO 2 and N 2 O 4 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 0.001369mol 2 /L 2 K=169 L/mol

13. N 2 + 3H 2  2NH 3 Write the K expression for the synthesis of ammonia.

14. Equilibrium Constant (K) [NH 3 ] 2 [N 2 ][H 2 ] 3 What will the units of K be in this example? K= L 2 /mol 2

15. Equilibrium Constant (K) At 300°C, the equilibrium concentrations are: [N 2 ] eq = 2.59M [H 2 ] eq =2.77M [NH 3 ] 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=3.3124mol 2 /L 2 55.05mol 4 /L 4 K=0.0602 L 2 /mol 2

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 NO 2 is reversed to be the decomposition of N 2 O 4 …

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 N 2 O 4 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) [N 2 O 4 ][NO 2 ] Initial0.7500 Change Eq.

24. Equilibrium Constant (K) [N 2 O 4 ][NO 2 ] Initial0.7500 Change-x+2x Eq.

25. Equilibrium Constant (K) [N 2 O 4 ][NO 2 ] Initial0.7500 Change-x+2x Eq.0.75 - x2x

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

27. Equilibrium Constant (K) 1.11 x 10 -3 = x2x2 0.0333= x

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: 1(0.0333) 1x  0.75 X 100 = 4.44% < 5%… assumption is good

31. Equilibrium Constant (K) [N 2 O 4 ][NO 2 ] Initial0.7500 Change-0.0333+2(0.0333) Eq.0.71670.0666

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

33. Equilibrium Constant (K) At 25°C, the initial concentration of NO 2 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 [NO 2 ][N 2 O 4 ] Initial0.5000 Change Final

36. Equilibrium Constant (K) Stoichiometry Chart [NO 2 ][N 2 O 4 ] Initial0.5000 Change-0.5+0.25 Final

37. Equilibrium Constant (K) Stoichiometry Chart [NO 2 ][N 2 O 4 ] Initial0.5000 Change-0.5+0.25 Final00.25

38. Equilibrium Constant (K) Equilibrium Chart [NO 2 ][N 2 O 4 ] Initial (=final) 00.25 Change Eq.

39. Equilibrium Constant (K) Equilibrium Chart [NO 2 ][N 2 O 4 ] Initial00.25 Change+2x-x Eq.

40. Equilibrium Constant (K) Equilibrium Chart [NO 2 ][N 2 O 4 ] Initial00.25 Change+2x-x Eq.2x0.25-x

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

42. Equilibrium Constant (K) 3.70 x 10 -4 = x2x2 0.0192= x

43. Equilibrium Constant (K) 5% rule test: 1(0.0192) 1x  0.25 X 100 = 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. Equilibrium Constant (K) [0.25 - x] [2 x] 2 169= [0.25 - x] 4x 2 169= 676x 2 = 0.25 - x

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

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

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

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

51. Equilibrium Constant (K) Equilibrium Chart [NO 2 ][N 2 O 4 ] Initial00.25 Change+2(0.0185)-0.0185 Eq.0.0370.2315

52. Equilibrium Constant (K) [0.2315] [0.037] 2 K= Check your answer: K= 169L/mol