1. Equilibrium

2. Reactions are reversible Z A + B C + D ( forward) Z C + D A + B (reverse) Z Initially there is only A and B so only the forward reaction is possible Z As C and D build up, the reverse reaction speeds up while the forward reaction slows down. Z Eventually the rates are equal

3. Reaction Rate Time Forward Reaction Reverse reaction Equilibrium

4. What is equal at Equilibrium? Z Rates are equal Z Concentrations are not. Z Rates are determined by concentrations and activation energy. Z The concentrations do not change at equilibrium. Z or if the reaction is verrrry slooooow.

5. Law of Mass Action Z For any reaction Z jA + kB lC + mD Z K = [C] l [D] m PRODUCTS power [A] j [B] k REACTANTS power Z K is called the equilibrium constant. Z is how we indicate a reversible reaction

6. Playing with K Z If we write the reaction in reverse. Z lC + mD jA + kB Z Then the new equilibrium constant is Z K = [A] j [B] k = 1/K [C] l [D] m

7. Playing with K Z If we multiply the equation by a constant Z njA + nkB nlC + nmD Z Then the equilibrium constant is Z K = [C] nl [D] nm ( [C] l [D] m ) n = K n [A] nj [B] nk = ([A] j [B] k ) n

8. The units for K Z Are determined by the various powers and units of concentrations. Z They depend on the reaction.

9. K is CONSTANT Z At any temperature. Z Temperature affects rate. Z The equilibrium concentrations dont have to be the same, only K. Z Equilibrium position is a set of concentrations at equilibrium. Z There are an unlimited number.

10. Equilibrium Constant One for each Temperature

11. Calculate K Z N 2 + 3H 2 2NH 3 Z Initial At Equilibrium Z [N 2 ] 0 =1.000 M [N 2 ] = 0.921M Z [H 2 ] 0 =1.000 M [H 2 ] = 0.763M Z [NH 3 ] 0 =0 M [NH 3 ] = 0.157M

12. Calculate K Z N 2 + 3H 2 2NH 3 Z Initial At Equilibrium Z [N 2 ] 0 = 0 M [N 2 ] = 0.399 M Z [H 2 ] 0 = 0 M [H 2 ] = 1.197 M Z [NH 3 ] 0 = 1.000 M [NH 3 ] = 0.203M Z K is the same no matter what the amount of starting materials

13. Equilibrium and Pressure Z Some reactions are gaseous Z PV = nRT Z P = (n/V)RT Z P = CRT Z C is a concentration in moles/Liter Z C = P/RT

14. Equilibrium and Pressure Z 2SO 2 (g) + O 2 (g) 2SO 3 (g) Z Kp = (P SO3 ) 2 (P SO2 ) 2 (P O2 ) Z K = [SO 3 ] 2 [SO 2 ] 2 [O 2 ]

15. Equilibrium and Pressure Z K = (P SO3 /RT) 2 (P SO2 /RT) 2 (P O2 /RT) Z K = (P SO3 ) 2 (1/RT) 2 (P SO2 ) 2 (P O2 ) (1/RT) 3 Z K = Kp (1/RT) 2 = Kp RT (1/RT) 3

16. General Equation Z jA + kB lC + mD Z K p = (P C ) l (P D ) m = (C C xRT ) l (C D xRT ) m (P A ) j (P B ) k (C A xRT ) j (C B xRT ) k Z K p = (C C ) l (C D ) m x(RT) l+m (C A ) j (C B ) k x(RT) j+k K p = K (RT) (l+m)-(j+k) = K (RT) n n= (l+m)-(j+k) = Change in moles of gas

17. Homogeneous Equilibria Z So far every example dealt with reactants and products where all were in the same phase. Z We can use K in terms of either concentration or pressure. Z Units depend on reaction.

18. Heterogeneous Equilibria Z If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesnt change. Z As long as they are not used up we can leave them out of the equilibrium expression. Z For example

19. For Example Z H 2 (g) + I 2 (s) 2HI(g) Z K = [HI] 2 [ H 2 ][ I 2 ] Z But the concentration of I 2 does not change. Z K [ I 2 ] = [HI] 2 = K [ H 2 ]

20. Write the equilibrium constant for the heterogeneous reaction

21. The Reaction Quotient Z Tells you the direction the reaction will go to reach equilibrium Z Calculated the same as the equilibrium constant, but for a system not at equilibrium Z Q = [Products] coefficient [Reactants] coefficient Z Compare value to equilibrium constant

22. What Q tells us Z If Q<K : Not enough products : Shift to right Z If Q>K : Too many products : Shift to left Z If Q=K system is at equilibrium

23. Example Z for the reaction Z 2NOCl(g) 2NO(g) + Cl 2 (g) Z K = 1.55 x 10 -5 M at 35ºC Z In an experiment 0.10 mol NOCl, 0.0010 mol NO(g) and 0.00010 mol Cl 2 are mixed in 2.0 L flask. Z Which direction will the reaction proceed to reach equilibrium?

24. Solving Equilibrium Problems Z Given the starting concentrations and one equilibrium concentration. Z Use stoichiometry to figure out other concentrations and K. Z Learn to create a table of initial and final conditions.

25. Z Consider the following reaction at 600ºC Z 2SO 2 (g) + O 2 (g) 2SO 3 (g) Z In a certain experiment 2.00 mol of SO 2, 1.50 mol of O 2 and 3.00 mol of SO 3 were placed in a 1.00 L flask. At equilibrium 3.50 mol of SO 3 were found to be present. Calculate Z The equilibrium concentrations of O 2 and SO 2, K and K P

26. Z Consider the same reaction at 600ºC 2SO 2 (g) + O 2 (g) 2SO 3 (g) Z In a different experiment.500 mol SO 2 and.350 mol SO 3 were placed in a 1.000 L container. When the system reaches equilibrium 0.045 mol of O 2 are present. Z Calculate the final concentrations of SO 2 and SO 3 and K

28. Solving Equilibrium Problems Type 1

29. What if youre not given equilibrium concentration? Z The size of K will determine what approach to take. Z First lets look at the case of a LARGE value of K ( >100). Z Allows us to make simplifying assumptions.

30. Example Z H 2 (g) + I 2 (g) 2HI(g) Z K = 7.1 x 10 2 at 25ºC Z Calculate the equilibrium concentrations if a 5.00 L container initially contains 15.8 g of H 2 294 g I 2. Z [ H 2 ] 0 = (15.8g/2.02)/5.00 L = 1.56 M Z [I 2 ] 0 = (294g/253.8)/5.00L = 0.232 M Z [HI] 0 = 0

31. Z Q= 0<K so more product will be formed. Z Set up table of initial, final and change in concentrations. Z Assumption since K is large- reaction will almost go to completion. Z Stoichiometry tells us I 2 is LR, it will be smallest at equilibrium let it be x

32. Z Choose X so it is small. Z For I 2 the change in X must be X-.232 M Z Final must = initial + change H 2 (g) I 2 (g) 2 HI(g) initial 1.56 M 0.232 M 0 M change final X

33. Z Using to stoichiometry we can find Z Change in H 2 = X-0.232 M Z Change in HI = -twice change in H 2 Z Change in HI = 0.464-2X H 2 (g) I 2 (g) 2 HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 M final X

34. Z Now we can determine the final concentrations by adding. H 2 (g) I 2 (g) 2 HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 M X-0.232 M 0.464-2X final X

35. Z Now plug these values into the equilibrium expression Z K = (0.464-2X) 2 = 7.1 x 10 2 (1.328+X)(X) H 2 (g) I 2 (g) 2 HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 M X-0.232 M 0.464-2X final 1.328+X X 0.464-2X

36. Why we chose X Z K = (0.464-2X) 2 = 7.1 x 10 2 (1.328+X)(X) Z Since X is going to be small, we can ignore it in relation to 0.464 and 1.328 Z So we can rewrite the equation Z 7.1 x 10 2 = (0.464) 2 (1.328)(X) Z Makes the algebra easy

37. Z When we solve for X we get 2.3 x 10 -4 Z So we can find the other concentrations Z I 2 = 2.3 x 10 -4 M Z H 2 = 1.328 M Z HI = 0.464 M H 2 (g) I 2 (g) 2 HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 M X-0.232 M 0.464-2X final 1.328+X X 0.464-2X

38. Checking the assumption Z The rule of thumb is that if the value of X is less than 5% of all the smallest concentrations, our assumption was valid. Z If not we would have had to use the quadratic equation Z More on this later. Z Our assumption was valid.

39. Practice Z For the reaction Cl 2 + O 2 2ClO(g) K = 156 Z In an experiment 0.100 mol ClO, 1.00 mol O 2 and 0.0100 mol Cl 2 are mixed in a 4.00 L flask. Z If the reaction is not at equilibrium, which way will it shift? Z Calculate the equilibrium concentrations.

40. At an elevated temperature, the reaction: has a value of K eq = 944. If 0.234 mol IBr is placed in a 1.00 L. flask and allowed to reach equilibrium, what is the equilibrium concentration in M. of I 2 ?

41. Type 2 Problems with small K K<.01

42. Process is the same Z Set up table of initial, change, and final concentrations. Z Choose X to be small. Z For this case it will be a product. Z For a small K the product concentration is small.

43. For example Z For the reaction 2NOCl 2NO +Cl 2 Z K= 1.6 x 10 -5 Z If 1.20 mol NOCl, 0.45 mol of NO and 0.87 mol Cl 2 are mixed in a 1 L container Z What are the equilibrium concentrations Z Q = [NO] 2 [Cl 2 ] = (0.45) 2 (0.87) = 0.15 M [NOCl] 2 (1.20) 2

44. Z Choose X to be small Z NO will be LR Z Choose NO to be X 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change Final

45. Z Figure out change in NO Z Change = final - initial Z change = X-0.45 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change Final X

46. Z Now figure out the other changes Z Use stoichiometry Z Change in Cl 2 is 1/2 change in NO Z Change in NOCl is - change in NO 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change X-.45 Final X

47. l Now we can determine final concentrations l Add 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change 0.45-X X-.45 0.5X -.225 Final X

48. l Now we can write equilibrium constant l K = (X) 2 (0.5X+0.645) (1.65-X) 2 l Now we can test our assumption X is small ignore it in + and - 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change 0.45-X X-.45 0.5X -.225 Final 1.65-X X 0.5 X +0.645

49. l K = (X) 2 (0.645) = 1.6 x 10 -5 (1.65) 2 l X= 8.2 x 10 -3 l Figure out final concentrations 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change 0.45-X X-.45 0.5X -.225 Final 1.65-X X 0.5 X +0.645

50. l [NOCl] = 1.64 l [Cl 2 ] = 0.649 l Check assumptions l.0082/.649 = 1.2 % OKAY!!! 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change 0.45-X X-.45 0.5X -.225 Final 1.65-X X 0.5 X +0.645

51. Practice Problem Z For the reaction 2ClO(g) Cl 2 (g) + O 2 (g) Z K = 6.4 x 10 -3 Z In an experiment 0.100 mol ClO(g), 1.00 mol O 2 and 1.00 x 10 -2 mol Cl 2 are mixed in a 4.00 L container. Z What are the equilibrium concentrations?

52. Type 3 Mid-range Ks.01<K<10

53. No Simplification Z Choose X to be small. Z Cant simplify so we will have to solve the quadratic (we hope) Z H 2 (g) + I 2 (g) 2HI(g) K=38.6 Z What is the equilibrium concentrations if 1.800 mol H 2, 1.600 mol I 2 and 2.600 mol HI are mixed in a 2.000 L container?

54. Problems Involving Pressure Z Solved exactly the same, with same rules for choosing X depending on K P Z For the reaction N 2 O 4 (g) 2NO 2 (g) K P =.131 atm. What are the equilibrium pressures if a flask initially contains 1.000 atm N 2 O 4 ?

55. Le Châteliers Principle Z If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress. Z 3 Types of stress a Concentration a Pressure a Temperature

56. Change amounts of reactants and/or products Z Adding product makes Q>K Z Removing reactant makes Q>K Z Adding reactant makes Q<K Z Removing product makes Q<K Z Determine the effect on Q, will tell you the direction of shift

57. Change Pressure Z By changing volume Z System will move in the direction that has the least moles of gas. Z Because partial pressures (and concentrations) change, a new equilibrium must be reached. Z System tries to minimize the moles of gas if volume is reduced Z And visa versa

58. Change in Pressure Z By adding an inert gas Z Partial pressures of reactants and product are not changed Z No effect on equilibrium position

59. Change in Temperature Z Affects the rates of both the forward and reverse reactions. Z Doesnt just change the equilibrium position, changes the equilibrium constant. Z The direction of the shift depends on whether it is exo- or endothermic

60. Exothermic DH<0 Z Releases heat Z Think of heat as a product Z Raising temperature push toward reactants. Z Shifts to left.

61. Endothermic DH>0 Z Produces heat Z Think of heat as a reactant Z Raising temperature push toward products. Z Shifts to right.