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Outline: 2/23/07

Outline: 2/23/07. Chem Dept Seminar – today@4pm CAPA 10 – deadline moved to Mon. Today: Finish Chapter 16. Chemical Equilibrium: LeChâtelier’s principle D G and K eq relationship. Types of Equilibrium Constants:. Lots of different names….

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Outline: 2/23/07

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  1. Outline: 2/23/07 • Chem Dept Seminar – today@4pm • CAPA 10 – deadline moved to Mon. • Today: Finish Chapter 16 • Chemical Equilibrium: LeChâtelier’s principle DG and Keq relationship

  2. Types of Equilibrium Constants: Lots of different names…. Keq, KH, Ksp, Ka , Kb, Kf , Kc, Kp… All the same idea: Extent of rxn… Keq is a number that describes the ratio of products/reactants…

  3. LeChâtelier’s Principle: • A system reacts to change in the direction that minimizes the change. • Change in concentrations • Change in pressures • Change in temperatures Demo

  4. LeChâtelier’s Principle Demo: Co(H2O)62+(aq) + 4Cl-CoCl4 2-(aq) + 6H2O Which way will the reaction shift when water is added? Toward reactants!!

  5. LeChâtelier’s Principle Demo: Co(H2O)62+(aq) + 4Cl-CoCl4 2-(aq) + 6H2O DH > 0 (endothermic) Co(H2O)62+(aq) +heatCoCl4 2-(aq) Which way will the reaction shift when it is heated? Toward products!!

  6. Practice problem: • PbCl2(s)Pb2+(aq) + 2 Cl-(aq) • Keq = [Pb2+][Cl-]2 • More PbCl2 is added: • More H2O is added: • Solid NaCl is added: • Solid KNO3 is added: • Nothing • More dissolvesà • ß More precip. • Nothing

  7. One last link to thermodynamics: DG = DGo+ RTlnQ At equilibrium: DG = 0 ; Q =Keq 0 = DGo+ RTlnKeq or-DGo = RTlnKeq or Keq= e-DGº/RT There is a relationship between DGo and Keq! (see CAPA-11)

  8. Example: Use the thermodynamic tables to find Keq for the following rxn: N2 + 3 H2 2 NH3 at 298 K DGo(kJ/mol) 0 0 -16.45 DGorxn = -32.9 kJ/mol Keq= e-DGo/RT = e32,900/(8.315*298) = 585000 Watch units!

  9. Example (non std. temp): Use the thermodynamic tables to find Keq for the following rxn: N2 + 3 H2 2 NH3 at 773 K DHo(kJ/mol) 0 0 -46.1 DSo(J/molK) 192 131 192 DHorxn = -92.2 kJ/mol DSorxn = -198.8 J/mol K DGTrxn = DHorxn- T DSorxn = 61.5 kJ Keq= e-DGT/RT = e-61,500/(8.315*773) = 7.010-5

  10. Example (non std. temp): Use the thermodynamic tables to find Keq for the following rxn: N2 + 3 H2 2 NH3 + heat • DHo(kJ/mol) 0 0 -46.1 • DSo(J/molK) 192 131 192 • DHorxn = -92.2 kJ/mol • DSorxn = -198.8 J/mol K • Exothermic reaction Keq@ 773K= 7.010-5 Keq@ 298K= 585000

  11. Example: Keq to Go You can also find Go given Keq N2 + 3 H2 2 NH3 at 773 K Keq= 7.010-5 DGo = - RTlnKeq DGo = - (8.314 J/k mol)(773 K) ln(7.0 10-5) =61.5 kJ/mol

  12. Let’s practice some more: • CAPA-11: problem #3 Some reaction… calculate Keq from DG… at 220 K! Keq= e-DGº/RT DGT = DHº-TDSº Keq= e-DHº/RT+DSº/R

  13. Let’s practice some more: 2A  B + C If 1.00 atm of A initially, and 0.24 atm of C at equilibrium…whats Keq? • CAPA-11: problem #7 0.24 atm C means 0.24 atm Bat equil. 0.24 atm C means 0.48 atm Adecays Keq= (0.24)(0.24)/ (1.00-.48)2

  14. Let’s practice some more: • CAPA-11: problems #9 & #10 A + B  C Kc = 710 calculate [C]/[B] at equilibrium if: 0.01 M A is added to 0.20 M B What must you do first? K is big: take reactants over to products

  15. x (0.19+x) (0.01-x) Equilibrium xis small… Let’s practice some more: A + B  C Kc = 710 • CAPA-11: problem #9 0.01 0.20 0.00 Initial -0.01 -0.01 +0.01 0.00 0.19 0.01 Initial +x+x-x Change K=710 = (0.01-x) / x (0.19+x) x= 7.41e-5

  16. Let’s practice some more: A + B  C Kc = 710 • CAPA-11: problem #10 7.41e-50.19 0.01Initial Dilute everything by 10.0… Does reaction quotient change? K=710 = (0.01) / (7.41e-5)(0.19) = (0.001) / (7.41e-6)(0.019) ? Q = 7100 Yes!

  17. (7.41e-6+x) (0.019+x) (0.001-x) Equil Let’s practice some more: Q = 7100 A + B  C Kc = 710 • CAPA-11: problem #9 7.41e-60.019 0.001Initial +x+x-x Change Q > Keq…which way does it go? Towards more reactants…

  18. Let’s test out those keypads…

  19. What is the equilibrium expression for: BaS(s) + 2 O2(g) BaSO4(s) • [O2]2 • [BaSO4]/[BaS][O2]2 • [BaS][O2]2/[BaSO4] • 1/[O2]2 • None of the above

  20. The Kp expression equals 400.0 for: CO(g) + H2O(g) CO2(g) + H2(g) If 4 atm each of CO & H2O are put into a container what’s the final pressure of CO ? • 0.04 atm • 0.2 atm • 5 atm • 80 atm • None of the above

  21. Have a great week-end!

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