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

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

n n We have defined a constant of the rxn: K eq = [products]/[reactants] Why is it so important?

Types of Equilibrium Constants: Salt Solubility Product: CuCl (s)  Cu + (aq) + Cl  (aq) K eq = [Cu + ][Cl  ] = K sp Acid-Base reactions: HA (aq) + H 2 O (aq)  H 3 O + (aq) + A  (aq) K eq = [H 3 O + ][A  ]/ [HA] = K a B (aq) + H 2 O (aq)  BH + (aq) + OH  (aq) K eq = [OH  ][BH  ]/ [B] = K b

Types of Equilibrium Constants: Gas-liquid : (e.g. vapor pressure) H 2 O ( )  H 2 O (g) K eq = p H 2 O Henry’s Law: K eq = K H Gas-solid : (e.g. vapor pressure) CO 2(s)  CO 2(g) K eq = p CO 2 Gas-aqueous: CO 2(g)  CO 2(aq) K eq = [CO 2 ]/ p CO 2 Demo K H

Types of Equilibrium Constants: Formation reactions: Ag + (aq) + 2NH 3(aq)  Ag(NH 3 ) 2 + (aq) K eq =[Ag(NH 3 ) 2 + ]/[Ag + ][NH 3 ] = K f Lots of different names…. K eq, K H, K sp, K a, K b, K f, K c, K p … All the same idea!  (aq)  (g)

Return to Worksheet #6  Do problems B & C & D

Worksheet #6 B. HOAc + H 2 O  H 3 O + + OAc  M M M M K eq = 1.8e  = [H 3 O + ][OAc  ] / [HOAc]  x x 0.20+x = (x)(0.20+x) / (0.050  x) Assume x is small… x = 4.5e  M

Worksheet #6 C. N 2 + 3H 2  2NH K eq = = [NH 3 ] 2 / [N 2 ][H 2 ]  x 0.25  x 2x = (2x) 2 / (0.15  x)(0.25  x) 3 4.8e  M x = 4.8e  M Is x small compared to 0.15 atm?

Worksheet #6: Last Problem init: 2 NO (g) + O 2(g)  2 NO 2(g) 5.0 atm 5.0 atm 0.0 atm K eq = 4.2  K eq = 4.2  = (p NO2 ) 2 /(p NO ) 2 (p O2 ) = (2x) 2 /(5  2x) 2 (5  x) ??? (5  2x) (5  x) +2x Given this K eq is x small? NO !

Worksheet #6: A new trick… init: 2 NO (g) + O 2(g)  2 NO 2(g) 5.0 atm 5.0 atm 0.0 atm K eq = 4.2  = (pNO 2 ) 2 /(pNO) 2 (pO 2 ) = (5  2x) 2 /(2x) 2 (2.5+x) ??? (5  5) (5  2.5) +5 Given this K eq is x small? 0.0 atm 2.5 atm 5.0 atm new init: (+2x) (2.5+x) (5  2x) YES !

Worksheet #6: Last Problem init: 2 NO (g) + O 2(g)  2 NO 2(g) 0.0 atm 2.5 atm 5.0 atm K eq = 4.2  = (pNO 2 ) 2 /(pNO) 2 (pO 2 ) = (5  2x) 2 /(2x) 2 (2.5+x) x = 7.7  10  (+2x) (2.5+x) (5  2x) = (5) 2 /(2x) 2 (2.5)