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Outline: 2/21/07

Outline: 2/21/07. Turn in Seminar Reports – to me Jaecker Applications – Chem Dept. Today: Chapter 16. Chemical Equilibrium: Types of K eq Manipulating/Calculating K eq LeChâtelier’s principle. Equilibrium Constant Rules:.

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Outline: 2/21/07

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  1. Outline: 2/21/07 • Turn in Seminar Reports – to me • Jaecker Applications– Chem Dept. • Today: Chapter 16 • Chemical Equilibrium: Types of Keq • Manipulating/Calculating Keq • LeChâtelier’s principle

  2. Equilibrium Constant Rules: Solids and (pure) liquids are left out of the Keq expression Reactions can be written in either direction at equilibrium Magnitude of Keq tells you about the extent of reaction Units of Keq are defined to be 1….

  3. We have defined a constant of the rxn: • Keq= [products]/[reactants] • Why is it so important?

  4. Types of Equilibrium Constants: Salt Solubility Product: CuCl(s) Cu+(aq) + Cl-(aq) Keq = [Cu+][Cl-] = Ksp Acid-Base reactions: HA(aq) + H2O(aq) H3O+(aq) + A-(aq) Keq = [H3O+][A-]/ [HA] = Ka B(aq) + H2O(aq) BH+(aq) + OH-(aq) Keq = [OH-][BH+]/ [B] = Kb

  5. Types of Equilibrium Constants: Gas-liquid : (e.g. vapor pressure) H2O(l) H2O(g) Keq = pH2O Gas-solid : (e.g. vapor pressure) CO2(s) CO2(g) Keq = pCO2 Gas-aqueous: CO2(g) CO2(aq) Keq = [CO2]/ pCO2 Henry’s Law: Keq = KH Demo KH

  6.  (aq) (g) • Types of Equilibrium Constants: Formation reactions: Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) Keq=[Ag(NH3)2+]/[Ag+][NH3] = Kf Lots of different names…. Keq, KH, Ksp, Ka , Kb, Kf , Kc, Kp… All the same idea!

  7. Return to Worksheet #6 • Do problems B & C & D

  8. Worksheet #6 • HOAc + H2O  H3O+ + OAc- 0.050 M0 0.20 M 0.050-xx0.20+x Keq = 1.8e-5 = [H3O+][OAc-] / [HOAc] = (x)(0.20+x)/ (0.050-x) Assume x is small… x = 4.5e-6 M

  9. Worksheet #6 • N2 + 3H2 2NH3 0.150.25 0 0.15-x0.25-3x2x Keq = 0.040 = [NH3]2 / [N2][H2]3 = (2x)2/ (0.15-x)(0.25-x)3 x = 4.8e-3 M Is x small compared to 0.15 atm?

  10. Worksheet #6: Last Problem 2 NO(g) + O2(g) 2 NO2(g) 5.0 atm 5.0 atm 0.0 atm init: (5-2x)(5-x) +2x Keq= 4.2  1012 Keq= 4.2  1012 = (pNO2)2/(pNO)2(pO2) = (2x)2/(5-2x)2(5-x) ??? Given this Keq is x small? NO !

  11. new init: 0.0 atm 2.5 atm 5.0 atm • Worksheet #6: A new trick… 2 NO(g) + O2(g) 2 NO2(g) 5.0 atm 5.0 atm 0.0 atm init: (5-5)(5-2.5) +5 (+2x)(2.5+x) (5-2x) Keq= 4.2  1012 = (pNO2)2/(pNO)2(pO2) = (5-2x)2/(2x)2(2.5+x) ??? YES ! Given this Keq is x small?

  12. Worksheet #6: Last Problem 2 NO(g) + O2(g) 2 NO2(g) 0.0 atm 2.5 atm 5.0 atm init: (+2x)(2.5+x) (5-2x) Keq= 4.2  1012 = (pNO2)2/(pNO)2(pO2) = (5-2x)2/(2x)2(2.5+x) = (5)2/(2x)2(2.5) Practice! x = 7.7  10-7

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