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Announcements

Announcements. Final Monday March 24 th at 6pm Cumulative 60% chapters 14 and 15, 40% rest of chapters Effective study techniques formulate exam questions (share with other students on canvas final study room)

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Announcements

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  1. Announcements • Final • Monday March 24th at 6pm • Cumulative • 60% chapters 14 and 15, 40% rest of chapters • Effective study techniques • formulate exam questions (share with other students on canvas final study room) • Teach it to someone (form a study group and take turns explaining different sections to each other) • Study guide is posted • Chapter 14 (mechanism to rate law) and worksheet study problems

  2. For extra help extra office hours schedule  • Monday 9-11am and 4-6pm • Tuesday 10am-12pm • Wednesday 10am-12am and 4-6pm • Thursday 10am-12pm • Monday (final day) 10-12am and 4-6pm • Please feel free to drop by my office during these hours 

  3. 15.7 Calculating equilibrium constant • Learning Objective: • To learn how to perform calculations involving equilibrium constants and concentrations or partial pressures of reactants.  SO2Cl2(g)  SO2(g) + Cl2(g)      

  4. Calculating KC from Equilibrium Concentrations • Two common types of calculations • Given equilibrium concentrations, calculate Kc • Given initial concentrations and one final concentration • Calculate equilibrium concentration of all other species • Then calculate Kc

  5. Calculating KC from Equilibrium Concentrations • 0.035 moles of SO2, 0.500 moles of SO2Cl2, and 0.080 moles of Cl2 are combined in an evacuated 5.00 L flask and heated to 100oC.  What is Q before the reaction begins?  Which direction will the reaction proceed in order to establish equilibrium? • SO2Cl2(g)  SO2(g) + Cl2(g)       Kc = 0.078 at 100oC 0.035 mole SO2/5.00 L = 0.070 M SO2 0.080 mole Cl2/5.00 L = 0.016 M Cl2 0.500 mole SO2Cl2/5.00 L = 0.100 M SO2Cl2

  6. Calculating KC from Equilibrium Concentrations SO2Cl2(g)  SO2(g) + Cl2(g)       Kc = 0.078 at 100oC Since K >Q, the reaction will proceed in the forward direction in order to increase the concentrations of both SO2and Cl2and decrease that of SO2Cl2until Q = K.

  7. Calculating KC Given Equilibrium Concentrations Ex. 3 N2O4(g)2NO2(g) • If you place 0.0350 mol N2O4 in 1 L flask at equilibrium, what is KC? If the concentrations at equilibrium are • [N2O4]eq = 0.0292 M • [NO2]eq = 0.0116 M KC = 4.61  10–3

  8. Your Turn! For the reaction: 2A(aq) + B(aq)3C(aq)the equilibrium concentrations are: A = 2.0 M, B = 1.0 M and C = 3.0 M. What is the expected value of Kc at this temperature? • 14 • 0.15 • 1.5 • 6.75

  9. Calculating KC Given Initial Concentrations and One Final Concentration Ex. 4 2SO2(g) + O2(g)2SO3(g) At 1000 K, 1.000 mol SO2 and 1.000 mol O2 are placed in a 1.000 L flask. At equilibrium 0.925 mol SO3 has formed. Calculate KC for this reaction. • First calculate concentrations of each • Initial • Equilibrium

  10. Ex. 4 (cont.) How to Solve: • Set up concentration table • Based on the following: • Changes in concentration must be in same ratio as coefficients of balanced equation • Set up table under balanced chemical equation • Initial concentrations • Controlled by person running experiment • Changes in concentrations • Controlled by stoichiometry of reaction • Equilibrium concentrations 2SO2(g) + O2(g) 2SO3(g) EquilibriumConcentration Change inConcentration InitialConcentration = –

  11. Ex. 4 (cont.) Concentration Table (ICE) –0.925 –0.462 +0.925 0.075 0.925 0.538 [SO3] at equilibrium = 0.925 M [SO2] consumed = amount of SO3 formed =0.925 M [O2] consumed = ½ amount SO3 formed = 0.925/2 = 0.462 M [SO2] at equilibrium = 1.000 – 0.975 = 0.075 [O2] at equilibrium = 1.00 – 0.462 = 0.538 M

  12. Ex. 4 Calculate Kc (cont.) • Finally calculate KC at 1000 K 0.075 0.925 0.538 Kc = 2.8 × 102 = 280

  13. H2 (g) + I2 (g) 2 HI (g) Equilibrium Calculations A closed system initially containing 1.000 x 10-3 M H2 and 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:

  14. H2 (g) + I2 (g) 2 HI (g) What Do We Know? ICEmethod:

  15. H2 (g) + I2 (g) 2 HI (g) [HI] Increases by 1.87 x 10-3 M

  16. H2 (g) + I2 (g) 2 HI (g) Stoichiometry tells us [H2] and [I2] decrease by half as much [I2] consumed =1/2 amount of HI formed = 1.87 x 10-3M/2= 9.35 x 10-4 [H2] consumed =1/2 amount of HI formed = 1.87 x 10-3M/2= 9.35 x 10-4

  17. H2 (g) + I2 (g) 2 HI (g) Stoichiometry tells us [H2] and [I2] decrease by half as much [I2] consumed =1/2 amount of HI formed = 1.87 x 10-3M/2= 9.35 x 10-4 [H2] consumed =1/2 amount of HI formed = 1.87 x 10-3M/2= 9.35 x 10-4

  18. H2 (g) + I2 (g) 2 HI (g) We can now calculate the equilibrium concentrations of all three compounds…

  19. (1.87 x 10-3)2 (6.5 x 10-5)(1.065 x 10-3) = = 51 …and, therefore, the equilibrium constant 6.5 x 10-5 1.065 x 10-3 1.87 x 10 -3

  20. Summary of Concentration Table • Used for most equilibrium calculations • Initial value in table (the units for all should be M) • [X]initial = those present when reaction prepared • No reaction occurs until everything is mixed • Values in last row of table are equilibrium concentrations • are the only values used to calculate Keq 3. The middle row Changes in concentrations always occur in same ratio as coefficients in balanced equation

  21. Recommended study problems 15.49, 15.50, 15.53, 15.54, 15.57, 15.58 15.59, 15.65, 15.70, 15.74

  22. Remember Beers law • Objective : We will explore an application of absorption spectroscopy using calibration curves and Beer’s Law. • Beer’s Law: • A = εlc A is absorbance ε is the molar absorptivity (in L/mol*cm), l is the path length (in cm), c is the concentration (in mol/L). A = nc Absorbance of a solution is directly proportional to the concentration of the solution and the length of solution the light has to pass through.

  23. Beers law • Collect absorbance data for the two unknown solutions. y=mx+b A = nc

  24. Keq of Bromothymol blue • Calculate Keq of Bromothymol Blue using a spectrophotometer and pH meter. • Indicators are typically weak acids or bases with complicated structures.

  25. Keq of Bromothymol blue • For simplicity, we represent a general indicator by the formula HIn- ,and its ionization in a solution by the equilibrium, Keep in mind that Bromothymol blue is blue when in the basic form (In2-) and yellow when in the acidic form (HIn-).

  26. Keq of Bromothymol blue

  27. Keq of Bromothymol blue • Collect absorbance data for the three basic and three acidic solutions y=mx+b A = nc

  28. Keq of Bromothymol blue [ In-] [HIn]

  29. Keq of Bromothymol blue Measure A [ In-] [HIn] • Measure pH

  30. pH and H3O+

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