1. Equilibrium Entry Task: Jan 8 th Tuesday At 1100 K, K p = 0.25 atm  1 for the reaction: 2 SO 2 (g) + O 2 (g) ↔ 2 SO 3 (g) What is the value of Kc at this temperature?

2. Equilibrium Agenda: Discuss Equilibrium constant Kp and Kc ws In-Class ICE table practice HW: Pre-Lab Determining Keq Constant

3. Equilibrium

4. 1. Calculate K p and K c for the following reaction….. CH 3 OH (g)  CO (g) + 2H 2 (g) Given the following equilibrium pressures at 25 o C: P CH3OH =6.10 x 10 -4 atm P CO = 0.387 atm P H2 = 1.34 atm Kc= 1.14 x 10 -3

5. Equilibrium 1. Calculate K p and K c for the following reaction….. CH 3 OH (g)  CO (g) + 2H 2 (g) Given the following equilibrium pressures at 25 o C: P CH3OH =6.10 x 10 -4 atm P CO = 0.387 atm P H2 = 1.34 atm Kp= Kc(RT) ∆n ∆n = 3-1 =2 Kp= (1.14 x 10 -3 )((0.0821)(298)) 2 Kp= 0.68

6. Equilibrium 2. A 127°C, Kc = 2.6 × 10  5 mol 2 /L 2 for the reaction: 2 NH 3 (g)  N 2 (g) + 3H 2 (g) Calculate K p at this temperature. Kp= Kc(RT) ∆n ∆n = 4-2 =2 Kp= (2.6 × 10  5 )((0.0821)(400)) 2 Kp= 2.8x10 -2

7. Equilibrium 3. At 1100 K, K p = 0.25 atm  1 for the reaction: 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) What is the value of Kc at this temperature? Kp= Kc(RT) ∆n ∆n = 2-3 =-1 0.25= (X)((0.0821)(1100)) -1 Kc= 23

8. Equilibrium 4. Phosphorus pentachloride dissociates on heating: PCl 5 (g)  PCl 3 (g) + Cl 2 (g) If Kc equals 3.28x10 -2 at 191°C, what is Kp at this temperature? Kp= Kc(RT) ∆n ∆n = 2-1 =1 Kp= (3.28x10 -2 )((0.0821)(464)) 1 Kp= 1.25

9. Equilibrium 5. Nitrogen dioxide dimerizes to form dinitrogen tetraoxide: 2 NO 2 (g)  N 2 O 4 (g) Calculate the value of K c, given that the gas phase equilibrium constant, K p, for the reaction is 1.3 × 10 3 at 273 K. (R = 0.08206 L·atm/mol·K) Kp= Kc(RT) ∆n ∆n = 1-2 = - 1 1.3 x10 -2 = (X)((0.08206)(273)) -1 Kc= 0.29

10. Equilibrium 6. An equilibrium mixture of SO 3, SO 2, and O 2 at 1000 K contains the gases at the following concentrations: [SO 3 ] = 0.41 M, [SO 2 ] = 0.032 M, and [O 2 ] = 0.59 M. What is the equilibrium constant (Kc) and the Kp value for the decomposition of SO 3 ? 2 SO 3 (g)  2 SO 2 (g) + O 2 (g) Kc= 3.59 x 10 -3

11. Equilibrium 6. An equilibrium mixture of SO 3, SO 2, and O 2 at 1000 K contains the gases at the following concentrations: [SO 3 ] = 0.41 M, [SO 2 ] = 0.032 M, and [O 2 ] = 0.59 M. What is the equilibrium constant (Kc) and the Kp value for the decomposition of SO 3 ? 2 SO 3 (g)  2 SO 2 (g) + O 2 (g) Kp= Kc(RT) ∆n ∆n = 3-2 =1 Kp= (3.59 x 10 -3 )((0.0821)(1000)) 1 Kp= 2.95 x 10 -1

12. Equilibrium

13. 15.4- Calculating Equilibrium Constant When a reaction has reached equilibrium, we often don’t know HOW the initial concentrations of the species have changed from the equilibrium concentrations. Or we were given JUST the initial concentrations, how can we determine the equilibrium concentrations?

14. Equilibrium

15. To look at all these variables at the same time we need to create an ICE table. I is initial concentration, C is the change in concentration E is the concentration at equilibrium.

16. Equilibrium There are two types of ICE tables Type 1 A.The initial or equilibrium concentration of some substances must be determined. B. Initial or equilibrium concentrations of some substances are given, but not both. Change is therefore treated as an unknown (x) C. The equilibrium constant is given.

17. Equilibrium There are two types of ICE tables Type 2 A. The equilibrium constant or concentration must be determined. B. Initial and equilibrium concentrations of at least one substance are given so that change can be calculated directly. C. All other initial and equilibrium concentrations of substances are determined directly from the table.

18. Equilibrium Walkthrough- (See pg. 571-2 Sample Exercise) A closed system initially containing 1.000 x 10 −3 M H 2 and 2.000 x 10 −3 M I 2 At 448  C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 −3 M. Calculate K c at 448  C for the reaction taking place, which is H 2 (g) + I 2 (g)  2 HI (g)

19. Equilibrium STEPS TO ICE ① Make Sure that all concentrations are in M- molarity!! (Done for you) ② Set up table- ICE (as you see it) and then species at top. ③ Place the known concentrations provided in the question into table. ④ Put in the CHANGE for HI (subtract equilibrium from initial)

20. Equilibrium ② Set up table- ICE (as you see it) and then species at top. [H 2 ], M[I 2 ], M[HI], M Initially Change At equilibrium

21. Equilibrium ③ Place the known concentrations provided in the question into table. [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change At equilibrium 1.87 x 10 -3

22. Equilibrium ④ Put in the CHANGE for HI (subtract equilibrium from initial) [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change+1.87 x 10 -3 At equilibrium 1.87 x 10 -3

23. Equilibrium ⑤ NOW we have to use the stoichiometry of the reaction to get the change of H 2 and I 2. Put a negative sign in front because they are reactants. [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change+1.87 x 10 -3 At equilibrium 1.87 x 10 -3 1.87 x 10 -3 mol L 1 mol H 2 2 mol of HI = 0.935 x 10 -3 Same goes for iodine H 2 (g) + I 2 (g)  2 HI (g)

24. Equilibrium ⑤ Stoichiometry tells us [H 2 ] and [I 2 ] decrease by half as much [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium 1.87 x 10 -3 The change MUST be in the negative because they are reactants!!

25. Equilibrium ⑥ Subtract the initial concentrations from the change which will provide the equilibrium value. [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium 1.87 x 10 -3 1.000 x10 -3 – 9.035 x10 -4 ) =6.5 x 10 -5 6.5 x 10 -5 1.065 x 10 -3 2.000 x10 -3 – 9.035 x10 -4 ) =1.065 x 10 -3

26. Equilibrium ⑦ Finally, provide the equilibrium expression for this reaction. Substitute the equilibrium values from chart into the expression to solve for Kc. = (1.87 x 10 -3 ) 2 (6.5 x 10 -5 )(1.065 x 10 -3 ) Kc =Kc = [HI] 2 [H 2 ] [I 2 ] Kc = 51

27. Equilibrium Let’s Try another Sulfur trioxide decomposes at high temperature in a sealed container: 2SO 3 (g)  2SO 2 (g) + O 2 (g). Initially, the vessel is charged at 1000 K with SO 3 (g) at a concentration of 6.09 x 10 -3 M. At equilibrium the SO 3 concentration is 2.44 x10 -3 M. Calculate the value of K p at 1000 K. INITIAL CHANGE EQUILIBRIUM 2SO 3  2SO 2 O2O2 6.09 x 10 -3 M0 M 2.44 x10 -3 M

28. Equilibrium Let’s Try another What information can we fill in with what we are given? INITIAL CHANGE EQUILIBRIUM 2SO 3  2SO 2 O2O2 6.09 x 10 -3 M0 M 2.44 x10 -3 M We can fill in the change of SO 3. 6.09 x 10 -3 M - 2.44 x10 -3 M = -3.65 x 10 -3 M -3.65 x 10 -3 M

29. Equilibrium Let’s Try another If SO 3 went down by -3.65 x 10 -3 M we have to use stoichiometry to find out the relationship the products have with the reactant. INITIAL CHANGE EQUILIBRIUM 2SO 3  2SO 2 O2O2 6.09 x 10 -3 M0 M 2.44 x10 -3 M 3.65 x 10 -3 M of SO 3 = +3.65 x 10 -3 M of SO 2 -3.65 x 10 -3 M 2 moles SO 2 2 mole of SO 3 +3.65 x 10 -3 M 3.65 x 10 -3 M of SO 3 = +1.83 x 10 -3 M of O 2 1 moles O 2 2 mole of SO 3 +1.83 x 10 -3 M

30. Equilibrium Let’s Try another Now subtract the initial from the change to get the equilibrium. INITIAL CHANGE EQUILIBRIUM 2SO 3  2SO 2 O2O2 6.09 x 10 -3 M0 M 2.44 x10 -3 M 0 - 3.65 x 10 -3 M = 3.65 x 10 -3 M -3.65 x 10 -3 M+3.65 x 10 -3 M 0 – 1.83 x 10 -3 M = 1.83 x 10 -3 M 1.83 x 10 -3 M +1.83 x 10 -3 M 3.65 x 10 -3 M

31. Equilibrium Let’s Try another Use the equilibrium concentrations to find Kc- plug n chug INITIAL CHANGE EQUILIBRIUM 2SO 3  2SO 2 O2O2 6.09 x 10 -3 M0 M 2.44 x10 -3 M -3.65 x 10 -3 M+3.65 x 10 -3 M 1.83 x 10 -3 M +1.83 x 10 -3 M 3.65 x 10 -3 M (1.33225 x10 -5 )(1.83 x10 -3 ) = 2.4380175 x10 -8 5.9536 x10 -6 Kc= 4.11 x 10 -3

32. Equilibrium In many situations we will know the value of the equilibrium constant and the initial concentrations of all species. We must then solve for the equilibrium concentrations. We have to treat them as variables or “x”

33. Equilibrium A chemist has a container of A 2 and B 2 and they react as given: A 2 (g) + B 2 (g)  2 AB (g) Kc = 9.0 at 100°C If 1.0 mole A 2 and 1.0 mole B 2 are placed in a 2.0 L container, what are the equilibrium concentrations of A 2, B 2, and AB? ① Convert to molarity!! 1 mol 2.0 L = 0.50 M

34. Equilibrium A chemist has a container of A 2 and B 2 and they react as given: A 2 (g) + B 2 (g)  2 AB (g) Kc = 9.0 at 100°C If 1.0 mole A 2 and 1.0 mole B 2 are placed in a 2.0 L container, what are the equilibrium concentrations of A 2, B 2, and AB? [A 2 ], M[B 2 ], M[AB], M Initially Change At equilibrium  0.50 M 0.0 M

35. Equilibrium [A 2 ], M[B 2 ], M[AB], M Initially Change At equilibrium  0.50 M 0.0 M Fill in the chart with variables!! You need to (for this ICE table) to factor in the stoich relationship) A 2 (g) + B 2 (g)  2AB (g) -x + 2x 0.50-x 2x

36. Equilibrium We have to work backwards!! Provide the Kc expression 9.0= (2x) 2 (0.50-x)(0.50-x) Kc =Kc = [AB] 2 [A 2 ] [B 2 ] [A 2 ], M[B 2 ], M[AB], M Initially0.50M 0 M Change At equilibrium -x  + 2x 0.50-x 2x

37. Equilibrium Get 0.50-x out by multiply on both sides Root both sides 9.0= (2x) 2 (0.50-x)(0.50-x) Or (0.50-x) 2 3.0= (2x) (0.50-x)

38. Equilibrium Get 0.50-x out by multiply on both sides 3.0= (2x) (0.50-x) 3.0(0.50-x)= 2x Multiply 3 through 1.5-3x= 2x Add 3x to both sides to get rid of -3x

39. Equilibrium 1.5-3x= 2x Add 3x to both sides to get rid of -3x 1.5 = 5x Divide both sides by 5 to get 5 out of there 0.3 = x SO back to the chart with our x value of 0.3

40. Equilibrium [A 2 ], M[B 2 ], M[AB], M Initially Change At equilibrium  0.50 M 0.0 M Now plug in our “x” value of 0.3 -x + 2x 2(0.3) = 0.6M 0.50-0.3 = 0.2M

41. Equilibrium Sample Exercise 15.11 A 1.000-L flask is filled with 1.000 mol of H 2 and 2.000 mol of I 2 at 448°C. The value of the equilibrium constant (Kc) is 50.5. H 2 (g) + I 2 (g)  2HI (g) What are the concentrations of H 2, I 2, and HI in the flask at equilibrium?

42. Equilibrium [H 2 ], M[I 2 ], M[HI], M Initially1.000M2.000 M0 M Change At equilibrium Fill in the chart with variables!! -x  + 2x 1.000-x2.000-x 2x You need to (for this ICE table) to factor in the stoich relationship) H 2 (g) + I 2 (g)  2HI (g)

43. Equilibrium [H 2 ], M[I 2 ], M[HI], M Initially1.000M2.000 M0 M Change At equilibrium Fill in the chart with variables!! -x  + 2x 1.000-x2.000-x 2x You need to (for this ICE table) to factor in the stoich relationship) H 2 (g) + I 2 (g)  2HI (g)

44. Equilibrium 50.5= (2x) 2 (1.000-x)(2.000-x) 50.5 = 4x 2 2-3x+x 2 Factor the denominator Get the denominator out of there multiply both sides 50.5 (2-3x+x 2 )= 4x 2 Multiply 51 though

45. Equilibrium 50.5 (2-3x+x 2 )= 4x 2 Multiply 51 though 101-151.5x+50.5x 2 =4x 2 Subtract 4x 2 from both sides 101-151.5x+46.5x 2 = 0 Quadratic equation C - Bx + Ax 2 =0 You have program your calculator OR do it the LONG WAY

46. Equilibrium 101-151.5x+46.5x 2 = 0 C - Bx + Ax 2 = X= -(-151.5) ± √151.5 2 – 4(46.5)(101) 2(46.5)

47. Equilibrium 101-151.5x+46.5x 2 = 0 C - Bx + Ax 2 = X= -(-151.5) ± √22952.25 –18786 93

48. Equilibrium 101-151.5x+46.5x 2 = 0 C - Bx + Ax 2 = X= -(-151.5) ± √4166.25 93

49. Equilibrium 101-151.5x+46.5x 2 = 0 C - Bx + Ax 2 = X= -(-151.5) + 64.5 93 = 2.32 X= -(-151.5) - 64.5 93 = 0.935

50. Equilibrium Which is the correct X= -(-151.5) + 64.5 93 = 2.32 X= -(-151.5) - 64.5 93 = 0.935 Substitute the value for X. If we get a negative concentration that value would NOT be the correct on.

51. Equilibrium [H 2 ], M[I 2 ], M[HI], M Initially1.000M2.000 M0 M Change At equilibrium -x  + 2x 1.000-x2.000-x 2x [H 2 ] 1.000- 2.32= -1.32 From Quadratic 1 NOT correct value!!

52. Equilibrium [H 2 ], M[I 2 ], M[HI], M Initially1.000M2.000 M0 M Change At equilibrium -0.935  2(0.935) 0.065M1.065 M 1.87 M [H 2 ] 1.000- 0.935= 0.065M From Quadratic 2 Correct value!!

53. Equilibrium [H 2 ], M[I 2 ], M[HI], M Initially1.000M2.000 M0 M Change At equilibrium -0.935  2(0.935) 0.065M1.065 M 1.87 M Plug the equilibrium concentrations into Kc expression see if its 50.5 Kc= 50.5

54. Equilibrium

57. HW: PreLab Determining Kc constant and ice tables ws

58. Equilibrium ICE TABLES

59. Equilibrium Sample Exercise 1 An equilibrium was established after 0.100 mol of hydrogen gas and 0.100 mol of iodine gas were added to an empty 1.00 L reaction vessel and heated to 700 K. The color intensity of the mixture changed from deep purple to a lighter purple color. At equilibrium, concentration of iodine was 0.0213 mol/L. Calculate the equilibrium concentrations of hydrogen gas and hydrogen iodide gas.

60. Equilibrium Sample Exercise 1-Plan a Strategy You will need to write the balanced chemical equation for the reaction, set up an ICE table, and enter the given data. Then, using the given data and your stoichiometric skills, you will find the change in the concentration of each species. This will involve finding change in [I 2 ] by subtraction and changes in [H 2 ] and [HI] by stoichiometry using the change in [I 2 ]. Update the ICE table as values are calculated. Calculate the equilibrium concentrations by addition or subtraction. Since H 2 and I 2 react to produce HI, their concentrations should be lower at equilibrium. Communicate the answers.

61. Equilibrium Sample Exercise 1- Step 1: Write the equation and build the ICE table.

62. Equilibrium Fill in information from question Initial Change Equilibriu m 0.100 M 0.0213 M 0.000 M

63. Equilibrium Sample Exercise 1- Step 2: Calculate the changes in [I 2 ], [H 2 ], and [HI]. Use the given change with the # moles

64. Equilibrium Fill in information from question Initial Change Equilibriu m 0.100 M 0.0213 M 0.0787 M 0.000 M

65. Equilibrium Sample Exercise 1- Step 2: Calculate the changes in [I 2 ], [H 2 ], and [HI]. Use mole ratios with the CHANGE in I 2

66. Equilibrium Fill in information from question Initial Change Equilibriu m 0.100 M 0.0213 M 0.0787 M 0.000 M

67. Equilibrium Sample Exercise 1- Step 2: Calculate the changes in [I 2 ], [H 2 ], and [HI]. Use mole ratios with the CHANGE in I 2

68. Equilibrium Fill in information from question Initial Change Equilibriu m 0.100 M 0.0213 M 0.0787 M 0.157 M 0.000 M

69. Equilibrium Sample Exercise 1- Step 3: Calculate the equilibrium concentrations. Subtract the initial from the change (reactants) Add the initial from the change (product) [H 2 ] Equilibrium = 0.100 M - 0.0787 M = 0.0213 M

70. Equilibrium Fill in information from question Initial Change Equilibriu m 0.100 M 0.0213 M 0.0787 M 0.157 M 0.0213 M 0.000 M

71. Equilibrium Sample Exercise 1- Step 3: Calculate the equilibrium concentrations. Subtract the initial from the change (reactants) Add the initial from the change (product) [HI] Equilibrium = 0.000 + 0.1574 = 0.1574 M

72. Equilibrium Fill in information from question Initial Change Equilibriu m 0.100 M 0.0213 M 0.0787 M 0.157 M 0.0213 M 0.000 M 0.157 M

73. Equilibrium Step 4: ANSWER At equilibrium the concentrations of hydrogen gas and hydrogen iodide gas are 0.0213 M and 0.157 M respectively

74. Equilibrium Things to note: You should see that the initial concentrations of the species placed in the vessel are greater than their equilibrium concentrations. That's because these species have been consumed to make the product. The product species have higher concentrations at equilibrium because initially they were not present in the vessel. All aqueous/gas species have non-zero concentrations at equilibrium. It is always a good idea to review the final numbers in an ICE table to ensure that the values make sense. For example, it would not make sense in the above example for the concentration of hydrogen gas to be higher at equilibrium than it was initially.

75. Equilibrium Sample Exercise 2 An equilibrium was established in a 1.00 L container when 0.120 mol of phosphorus pentachloride gas was decomposed at 500 K. At equilibrium, the concentration of chlorine gas was 0.0540 mol/L. Calculate the equilibrium constant for this system and state a conclusion about the position of the equilibrium

76. Equilibrium Sample Exercise 2- Step 1: Write the equation and build the ICE table.

77. Equilibrium Fill in information from question Initial Change Equilibriu m 0.120 M0.000 M 0.0540 M 0.000 M

78. Equilibrium Sample Exercise 2- Step 2: Calculate the changes in given species. Use the given change with the # moles

79. Equilibrium Fill in information from question Initial Change Equilibriu m 0.120 M0.000 M 0.0540 M 0.000 M 0.0540 M

80. Equilibrium Sample Exercise 2- Step 2: Calculate the changes in given species. Now we have 1 for the change can use stoich to figure out the other changes

81. Equilibrium Fill in information from question Initial Change Equilibriu m 0.120 M0.000 M 0.0540 M 0.000 M 0.0540 M

82. Equilibrium Sample Exercise 2- Step 2: Calculate the changes in given species. Now we have 1 for the change can use stoich to figure out the other changes

83. Equilibrium Fill in information from question Initial Change Equilibriu m 0.120 M0.000 M 0.0540 M 0.000 M 0.0540 M

84. Equilibrium Sample Exercise 2- Step 3: Calculate the equilibrium concentrations. Subtract the initial from the change (reactants) Add the initial from the change (product) [PCl 5 ] Equilibrium = 0.120 M - 0.0540 M = 0.066 M

85. Equilibrium Fill in information from question Initial Change Equilibriu m 0.120 M0.000 M 0.0540 M 0.000 M 0.0540 M 0.0066 M

86. Equilibrium Sample Exercise 2- Step 3: Calculate the equilibrium concentrations. Subtract the initial from the change (reactants) Add the initial from the change (product) [PCl 3 ] Equilibrium = 0.000 + 0.0540 = 0.0540 M

87. Equilibrium Fill in information from question Initial Change Equilibriu m 0.120 M0.000 M 0.0540 M 0.000 M 0.0540 M 0.0066 M0.0540 M

88. Equilibrium Sample Exercise 2- Step 4: Write the K expression and solve.

89. Equilibrium Sample Exercise 2- Calculate the equilibrium constant for this system and state a conclusion about the position of the equilibrium The equilibrium constant for the decomposition of phosphorus pentachloride at 500 K is 4.4 x 10 -2. The equilibrium position lies to the left, or in other words, the reactant is favored.