1. Le Châtelier’s principle

2. The significance of Kc values If Kc is small (0.001 or lower), [products] must be small, thus forward reaction is weak If Kc is large (1000 or more), [products] must be large, thus forward reaction is strong [Products] Kc = [Reactants] Reactants  Products If Kc is about 1, then reactants and products are about equal but not exactly since they may be raise to different exponents

3. Stresses to equilibria Changes in reactant or product concentrations is one type of “stress” on an equilibrium Other stresses are temperature, and pressure.

4. The response of equilibria to these stresses is explained by Le Chatelier’s principle: If an equilibrium in a system is upset, the system will tend to react in a direction that will reestablish equilibrium Thus we have: 1) Equilibrium, 2) Disturbance of equilibrium, 3) Shift to restore equilibrium Le Chatelier’s principle predicts how an equilibrium will shift (

5. N 2 + 3H 2  2NH 3 + 92 kJ Summary of Le Chatelier’s principle E.g. N 2 + 3H 2  2 NH 3 + 92 kJ Pressure (due to decreased volume): increase in pressure favors side with fewer molecules Amounts of products and reactants: equilibrium shifts to compensate  N 2  H 2 Temperature: equilibrium shifts to compensate:  Heat shift right shift left

6. Part 1: changes in reactant or product concentrations Cu +2 + 4NH 3 Cu(NH 3 ) 4 +2 + 4H 2 O Ni +2 + 4NH 3 Ni(NH 3 ) 6 +2 + 6H 2 O NH 3 + H + (NH4)

7. Part A: copper and nickel ions Do not do Nickel ions Question 4 the answer is pale green Question 5 the answer is Violet Question 6 the answer is pale green

8. Part B: Cobalt ions Do not do Co +2 + 4Cl - CoCl 4 -2 + 6H 2 O Question 8 the answer is pale pink Question 9 the answer is deep blue Question 10 the answer is pale pink

9. Part II. Equilibria involving sparingly soluble salts Ag + + CO 3 -2 Ag 2 CO 3 -2 2H + + CO 3 -2 H 2 O + CO 2

10. Part II. Equilibria involving sparingly soluble salts Ag + + Cl - AgCl Ag + + 2NH 3 Ag(NH 3 ) 2 + heat NH 3 + H + (NH4)

11. Part III. Temp on Equilibria Heat + CoCl 2 + 6H 2 O Co(H 2 O) 6 + 4Cl