2. Previously in Chem 104: Kinetics TODAY A new chapter, a new idea? A moment of review: Big Ideas in Chemistry 1. Structure determines Behavior 2. Equilibrium and

3. TODAY A new chapter: Equilibrium Writing equilibrium expressions Disturbing Equilibrium; observing LeChatelier’s Principle Calculating equilibrium constants, K

4. At the end of Kinetics Chapter, this was introduced: NO + NO N2O2N2O2 k1k1 k -1 Dynamic applet

5. At the end of Kinetics Chapter, this was introduced: NO + NO N2O2N2O2 k1k1 k -1 And; k -1 k1k1 K eq Backward rate Forward rate or

6. What does the value of Keq tell us? NO + NO N2O2N2O2 k1k1 k -1 If: 10 -13 M sec -1 10 3 M sec -1 Slow backward rate Fast forward rate like Reaction “lies” towards products, K eq is LARGE

7. What does the value of Keq tell us? NO + NO N2O2N2O2 k1k1 k -1 If: 10 -3 M sec -1 10 -13 M sec -1 Fast backward rate Slow forward rate like Reaction “lies” towards reactants, K eq is small: We say reaction doesn’t “go”.

8. Are rate constants the only way to get K eq ? NO + NON2O2N2O2 For: K eq [reagents] [products] NO ! K eq [NO] 2 [N 2 O 2 ]

9. More generally K eq is defined as: aA + bBcC + dD For: K eq [reagents] [products] That’s all you need know in Chapter 16. K eq [A] a [B] b [C] c [D] d

10. What is this Copper experiment about? Closely followed by “ a system at equilibrium when disturbed will adjust to remove or minimize the effect of the disturbance”

11. The series of reactions observed: 3. [Cu(H 2 O) 6 ] 2+ + Cl-[CuCl(H 2 O) 5 ] 2+ + H 2 O 1. CuCl 2 (H 2 O) 2 (s) + 3H 2 O[CuCl(H 2 O) 5 ] + + Cl- 2. [CuCl(H 2 O) 5 ] + + H 2 O[Cu(H 2 O) 6 ] 2+ + Cl-

12. What is first equilibrium expression: [Cu(H 2 O) 6 ] 2+ + Cl- K eq [reagents] [products] K eq [Cu(H 2 O) 6 2+ ][Cl] [CuCl(H 2 O) 5 + ][H 2 O] [CuCl(H 2 O) 5 ] + + H 2 O K eq [Cu(H 2 O) 6 2+ ][Cl] [CuCl(H 2 O) 5 + ] But, pure liquids don’t appear in K eq expression:

13. The series of reactions observed: 3. [Cu(H 2 O) 6 ] 2+ + Cl-[CuCl(H 2 O) 5 ] 2+ + H 2 O K eq [Cu(H 2 O) 6 2+ ][Cl] [CuCl(H 2 O) 5 2+ ] 1. CuCl 2 (H 2 O) 2 (s) + 3H 2 O[CuCl(H 2 O) 5 ] + + Cl- 2. [CuCl(H 2 O) 5 ] + + H 2 O[Cu(H 2 O) 6 ] 2+ + Cl- K eq 0.40M x 7.1 M 0.80 M 0.28, NO UNITS!

14. Combining reactions equals multiplying K’s K 1 eq x K 2 eq (a) CuCl 2 (H 2 O) 2 (s) + 3H 2 O[CuCl(H 2 O) 5 ] + + Cl- K 1 eq Net CuCl 2 (H 2 O) 2 (s) + 4H 2 O[Cu(H 2 O) 6 ] 2+ + 2 Cl- K net eq (b) [CuCl(H 2 O) 5 ] + + H 2 O[Cu(H 2 O) 6 ] 2+ + Cl- K 2 eq [Cu(H 2 O) 6 2+ ][Cl] [CuCl(H 2 O) 5 2+ ] CuCl 2 (H 2 O) 2 (s) [CuCl(H 2 O) 5 2+ [Cl] [Cu(H 2 O) 6 2+ ][Cl] 2 CuCl 2 (H 2 O) 2 (s) K net eq [Cu(H 2 O) 6 2+ ][Cl] 2 Pure solids don’t appear in Keq

15. What was this Copper experiment about? Closely followed by “ a system at equilibrium when disturbed will adjust to remove or minimize the effect of the disturbance”