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1. Entry Task: Jan 8th Tuesday At 1100 K, Kp = 0.25 atm1 for the reaction: 2 SO2(g) + O2(g) ↔ 2 SO3(g) What is the value of Kcat this temperature?

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

3. 1. Calculate Kp and Kc for the following reaction….. CH3OH (g) CO (g) + 2H2 (g) Given the following equilibrium pressures at 25oC: P CH3OH=6.10 x 10-4atm PCO= 0.387 atm PH2= 1.34 atm Kc= 1.14 x 10-3

4. 1. Calculate Kp and Kc for the following reaction….. CH3OH (g) CO (g) + 2H2 (g) Given the following equilibrium pressures at 25oC: P CH3OH=6.10 x 10-4atm PCO= 0.387 atm PH2= 1.34 atm Kp= Kc(RT)∆n ∆n = 3-1 =2 Kp= (1.14 x 10-3)((0.0821)(298))2 Kp= 0.68

5. 2. A 127°C, Kc = 2.6 × 105 mol2/L2 for the reaction: 2 NH3(g)  N2(g) + 3H2(g) Calculate Kp at this temperature. Kp= Kc(RT)∆n ∆n = 4-2 =2 Kp= (2.6 × 105)((0.0821)(400))2 Kp= 2.8x10-2

6. 3. At 1100 K, Kp = 0.25 atm1 for the reaction: 2 SO2(g) + O2(g)  2 SO3(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

7. 4. Phosphorus pentachloride dissociates on heating: PCl5(g)  PCl3(g) + Cl2(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

8. 5. Nitrogen dioxide dimerizes to form dinitrogentetraoxide: 2 NO2(g) N2O4(g) Calculate the value of Kc, given that the gas phase equilibrium constant, Kp, for the reaction is 1.3 × 103 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

9. 6. An equilibrium mixture of SO3, SO2, and O2 at 1000 K contains the gases at the following concentrations: [SO3] = 0.41 M, [SO2] = 0.032 M, and [O2] = 0.59 M. What is the equilibrium constant (Kc) and the Kp value for the decomposition of SO3? 2 SO3(g) 2 SO2(g) + O2(g) Kc= 3.59 x 10-3

10. 6. An equilibrium mixture of SO3, SO2, and O2 at 1000 K contains the gases at the following concentrations: [SO3] = 0.41 M, [SO2] = 0.032 M, and [O2] = 0.59 M. What is the equilibrium constant (Kc) and the Kp value for the decomposition of SO3? 2 SO3(g) 2 SO2(g) + O2(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

11. 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?

12. To look at all these variables at the same time we need to create an ICE table. Iis initial concentration, Cis the change in concentration Eis the concentration at equilibrium.

13. There are two types of ICE tables Type 1 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.

14. 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.

15. H2 (g) + I2 (g) 2 HI (g) Walkthrough-(See pg. 571-2 Sample Exercise) A closed system initially containing 1.000 x 10−3 M H2and 2.000 x 10−3 M I2 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 Kc at 448C for the reaction taking place, which is

16. 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. ③Place the known concentrations provided in the question into table. ④Put in the CHANGE for HI (subtract equilibrium from initial)

17. H2 (g) + I2 (g) 2 HI (g) ⑤NOW we have to use the stoichiometry of the reaction to get the change of H2 and I2. Put a negative sign in front because they are reactants. 1.87 x 10-3mol L 1 mol H2 2 mol of HI = 0.935 x 10-3 Same goes for iodine

18. ⑤ Stoichiometry tells us [H2] and [I2]decrease by half as much The change MUST be in the negative because they are reactants!!

19. ⑥ Subtract the initial concentrations from the change which will provide the equilibrium value. • 6.5 x 10-5 • 1.065 x 10-3 1.000 x10-3 – 9.035 x10-4) =6.5 x 10-5 2.000 x10-3 – 9.035 x10-4) =1.065 x 10-3

20. (1.87 x 10-3)2 (6.5 x 10-5)(1.065 x 10-3) = ⑦ Finally, provide the equilibrium expression for this reaction. Substitute the equilibrium values from chart into the expression to solve for Kc. [HI]2 [H2] [I2] Kc= Kc = 51

21. Let’s Try another Sulfur trioxide decomposes at high temperature in a sealed container: 2SO3(g)  2SO2(g) + O2(g). Initially, the vessel is charged at 1000 K with SO3(g) at a concentration of 6.09 x 10-3 M. At equilibrium the SO3 concentration is 2.44 x10-3M. Calculate the value of Kpat 1000 K. O2 2SO2 2SO3  0 M 6.09 x 10-3 M 0 M 2.44 x10-3M

22. Let’s Try another O2 2SO2 2SO3  0 M 6.09 x 10-3 M 0 M -3.65 x 10-3 M 2.44 x10-3M What information can we fill in with what we are given? We can fill in the change of SO3. 6.09 x 10-3 M - 2.44 x10-3M = -3.65 x 10-3 M

23. Let’s Try another O2 2SO2 2SO3  0 M 6.09 x 10-3 M 0 M -3.65 x 10-3M +3.65 x 10-3 M +1.83 x 10-3 M 2.44 x10-3M If SO3 went down by -3.65 x 10-3M we have to use stoichiometry to find out the relationship the products have with the reactant. 2 moles SO2 2 mole of SO3 = +3.65 x 10-3 M of SO2 3.65 x 10-3M of SO3 1 moles O2 2 mole of SO3 = +1.83 x 10-3 M of O2 3.65 x 10-3M of SO3

24. Let’s Try another O2 2SO2 2SO3  0 M 6.09 x 10-3 M 0 M -3.65 x 10-3M +3.65 x 10-3M +1.83 x 10-3 M 3.65 x 10-3M 2.44 x10-3M 1.83 x 10-3 M Now subtract the initial from the change to get the equilibrium. 0 - 3.65 x 10-3M = 3.65 x 10-3M 0 – 1.83 x 10-3M = 1.83 x 10-3 M

25. Let’s Try another O2 2SO2 2SO3  0 M 6.09 x 10-3 M 0 M -3.65 x 10-3M +3.65 x 10-3M +1.83 x 10-3 M 3.65 x 10-3M 2.44 x10-3M 1.83 x 10-3 M Use the equilibrium concentrations to find Kc- plug n chug (1.33225 x10-5)(1.83 x10-3) = 2.4380175 x10-8 5.9536 x10-6 Kc= 4.11 x 10-3

26. 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”

27. A chemist has a container of A2 and B2 and they react as given: A2 (g) + B2 (g)  2 AB (g) Kc= 9.0 at 100°CIf 1.0 mole A2 and 1.0 mole B2 are placed in a 2.0 L container, what are the equilibrium concentrations of A2, B2, and AB? ① Convert to molarity!! 1 mol 2.0 L = 0.50 M

28. A chemist has a container of A2 and B2 and they react as given: A2 (g) + B2 (g)  2 AB (g) Kc= 9.0 at 100°CIf 1.0 mole A2 and 1.0 mole B2 are placed in a 2.0 L container, what are the equilibrium concentrations of A2, B2, and AB?  0.50 M 0.50 M 0.0 M

29. 0.50 M 0.50 M 0.0 M -x -x + 2x 2x 0.50-x 0.50-x Fill in the chart with variables!! A2(g) + B2(g) 2AB (g) You need to (for this ICE table) to factor in the stoich relationship)

30. (2x)2 (0.50-x)(0.50-x) 9.0=  -x -x + 2x 2x 0.50-x 0.50-x We have to work backwards!! [AB]2 [A2] [B2] Kc= Provide the Kc expression

31. (2x)2 (0.50-x)(0.50-x) Or (0.50-x)2 (2x) (0.50-x) 9.0= 3.0= Root both sides Get 0.50-x out by multiply on both sides

32. (2x) (0.50-x) 3.0= 2x 3.0(0.50-x)= Get 0.50-x out by multiply on both sides Multiply 3 through 1.5-3x= 2x Add 3x to both sides to get rid of -3x

33. 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 SO back to the chart with our x value of 0.3 0.3 = x

34. 0.50 M 0.50 M 0.0 M -x -x + 2x 0.50-0.3 = 0.2M 0.50-0.3 = 0.2M 2(0.3) = 0.6M Now plug in our “x” value of 0.3

35. Sample Exercise 15.11 A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of the equilibrium constant (Kc) is 50.5. H2(g) + I2(g) 2HI (g) What are the concentrations of H2, I2, and HI in the flask at equilibrium?

36. -x -x + 2x 2x 2.000-x 1.000-x Fill in the chart with variables!! H2(g) + I2(g)  2HI (g) You need to (for this ICE table) to factor in the stoich relationship)

37. -x -x + 2x 2x 2.000-x 1.000-x Fill in the chart with variables!! H2(g) + I2(g)  2HI (g) You need to (for this ICE table) to factor in the stoich relationship)

38. (2x)2 (1.000-x)(2.000-x) 4x2 2-3x+x2 50.5= 50.5 = 50.5 (2-3x+x2)= 4x2 Factor the denominator Get the denominator out of there multiply both sides Multiply 51 though

39. 50.5 (2-3x+x2)= 101-151.5x+50.5x2= 101-151.5x+46.5x2= C - Bx + Ax2 = 0 4x2 4x2 0 Multiply 51 though Subtract 4x2 from both sides Quadratic equation You have program your calculator OR do it the LONG WAY

40. 101-151.5x+46.5x2= 0 C - Bx + Ax2 = X= -(-151.5)±√151.52 – 4(46.5)(101) 2(46.5)

41. 101-151.5x+46.5x2= 0 C - Bx + Ax2 = X= -(-151.5)±√22952.25 –18786 93

42. 101-151.5x+46.5x2= 0 C - Bx + Ax2 = X= -(-151.5)±√4166.25 93

43. 101-151.5x+46.5x2= 0 C - Bx + Ax2 = X= -(-151.5)+ 64.5 = 2.32 93 X= -(-151.5)- 64.5 = 0.935 93

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