Outline:2/19/07 è è Today: Chapter 16 Ù Chemical Equilibrium Ù The Equilibrium Constant (K eq ) Ù Manipulating/Calculating K eq è Pick up Exam #1 - from me è Pick up CAPA #10 & 11 - outside è Seminar – noon è Jaecker Applications – Chem Dept.

Exam #1 è Average ~74.7 è High: 99,… Well done!

Exam #1 è What to do if…. Go over test = do a post mortem Remember it’s only 20%...or less! Change something… Come visit…

  2 NO 2(g)  N 2 O 4(g) Rate formation = k f [NO 2 ] 2   N 2 O 4(g)  2 NO 2(g) Rate decomposition = k d [N 2 O 4 ] At equilibrium: Rate f = Rate d k f [ NO 2 ] 2 = k d [ N 2 O 4 ] Chapter 16: Equilibrium

Define an “equilibrium constant”: K eq = k f /k d = [ N 2 O 4 ]/[ NO 2 ] 2 nomenclature: [concentrations] or p   becomes    Rxn quotient: Q = [products]/[reactants]  e.g. If Q = K eq system is at equilibrium  If Q > K eq products  reactants  If Q < K eq reactants  products

The Equilibrium Constant For a general reaction the equilibrium constant expression for everything in solution is where K eq is the equilibrium constant.

The Equilibrium Constant An equilibrium can be approached from any direction. Example: has

In the reverse direction: The Equilibrium Constant

Equilibrium Constant Rules: Magnitude of K eq tells you about preferred direction of reaction K eq = or K eq = 4.6  10  8 more products (at equilibrium) more reactants

Equilibrium Constant Rules: Reactions can be written in either direction at equilibrium K eq = 1/ K ’ eq Reactions can be written as the sum of other reactions (Hess’ law) K eq = K ’ eq K ” eq

A specific example (chem 113):  AgCl (s)  Ag + + Cl   eq = 1.8  10  10  Ag NH 3  Ag(NH 3 ) 2 +   eq = 1.6  10 7   eq = 2.9  10  3  AgCl (s) + 2 NH 3  Ag(NH 3 ) Cl    eq =

Equilibrium Constant Rules: Solids and (pure) liquids are left out of the K eq expression Example: H 3 PO 4(aq) + 3 H 2 O (l)  PO 3  4(aq) + 3H 3 O + (aq) K eq = [PO 3  4 ] [H 3 O + ] 3 / [H 3 PO 4 ]

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

Let’s Practice: K eq = (p NO ) 2 (p O2 ) / (p NO2 ) 2 Worksheet #6 (A) = (0.0015) 2  / (0.1) 2 = 1.15e-5

  1  10  3 1  10  3 0 Initial    x  x  x Change (1  10  3  x )(1  10  3  x ) +x Equil. Starting from initial concentrations:  Mix: 1  10  3 Fe 3+ and 1  10  3 SCN  K eq = 142 What are the final concentrations?  Fe 3+ + SCN   Fe(SCN) 2+

  K eq = [Fe(SCN) 2+ ]/[Fe 3+ ][SCN  ]   142 = x / (1  10  3  x ) (1  10  3  x ) (a quadratic equation in x ) n Initial concentrations:  Mix: 1  10  3 Fe 3+ and 1  10  3 SCN  What are the final concentrations?  x = 1.11  10  4

Return to Worksheet #6  Do problems B & C Practice these! Finish Friday