1. Previously in Chem104 A new chapter: Equilibrium Writing equilibrium expressions observing LeChatelier’s Principle Calculating equilibrium constants, K TODAY A new wrinkle on Equilibrium: Q Q, from the Q continuum, Star Trek,The Next Generation No, not that Q..

2. TODAY A new chapter: Equilibrium Writing equilibrium expressions observing LeChatelier’s Principle Calculating equilibrium constants, K TODAY A quick recap of Monday’s concepts Magnitudes of K eq Q: the reaction quotient Q: how to use it

3. A quick recap on equilibrium expressions: K eq [reagents] [products] [Cu(H 2 O) 6 ] 2+ + Cl-[CuCl(H 2 O) 5 ] + + H 2 O For this reaction: But if reaction goes BOTH to right and to left, which are reagents and which are products? By “convention” reagents are species to the left of arrow products are species to the right of arrow

4. A quick recap on equilibrium behavior: [Cu(H 2 O) 6 ] 2+ + Cl-[CuCl(H 2 O) 5 ] + + H 2 O You watched: You added excess Cl- : [Cu(H 2 O) 6 ] 2+ + Cl- [CuCl(H 2 O) 5 ] + + H 2 O You added excess H 2 O: [Cu(H 2 O) 6 ] 2+ + Cl- [CuCl(H 2 O) 5 ] + + H 2 O

5. A quick recap on equilibrium expressions: [Cu(H 2 O) 6 ] 2+ + Cl- K eq [reagents] [products] [CuCl(H 2 O) 5 ] + + H 2 O K eq [Cu(H 2 O) 6 2+ ][Cl] [CuCl(H 2 O) 5 + ] Pure liquids and solids don’t appear in K eq expression: You watched: You wrote: 0.28 When [CuCl(H 2 O) 5 + ]= 0.8 M, [Cu(H 2 O) 6 2+ ]= 0.4 M, [Cl] =7.1M You know:

6. Are Your Eyes Misleading You? What is in the graduated cylinder? Abs wavelength 700 nm 400 nm [Cu(H 2 O) 6 2+ ] [CuCl(H 2 O) 5 + ] Visible electronic spectra mixture can appear GREEN

7. Are Your Eyes Misleading You? What is in the graduated cylinder? K eq [Cu(H 2 O) 6 2+ ][Cl] [CuCl(H 2 O) 5 + ] Recall equilibrium concentrations: 0.28 0.8 M 0.4 M x 7.1M Abs wavelength 700 nm 400 nm [Cu(H 2 O) 6 2+ ] [CuCl(H 2 O) 5 + ] mixture can appear GREEN Visible electronic spectra If K eq ~ 1, product concentrations are similar to reagent concentrations

8. How large must K eq be for reaction to be “complete”? Consider these reactions: K eq 2.0 x 10 8 [Ni(NH 3 ) 6 ] 2+ + 6H 2 O [Ni(en) 3 ] 2+ + 6 NH 3 [Ni(NH 3 ) 6 ] 2+ + 3 “en” K eq 7.3 x 10 9

9. How large must K eq be for reaction to be “complete”? Consider this reaction: K eq 2.0 x 10 8 [Ni(NH 3 ) 6 ] 2+ + 6H 2 O

10. What is K eq for this reaction: ? K 1 = 2.0 x 10 8 [Ni(NH 3 ) 6 ] 2+ + 6H 2 O [Ni(en) 3 ] 2+ + 6 NH 3 [Ni(NH 3 ) 6 ] 2+ + 3 “en”K 2 = 7.3 x 10 9 [Ni(en) 3 ] 2+ + 6 H 2 O K = K 1 x K 2 = (7.3 x 10 9 )(2.0 x 10 8 ) = 1.5 x 10 18 K = K 1 x K 2

11. This K eq seems huge! What is ratio of rgt. Ni to prdt. Ni species [Ni(en) 3 ] 2+ + 6 H 2 O K = 1.5 x 10 18

12. What happens if the concentrations are equal: [Cu(H 2 O) 6 ] 2+ + Cl-[CuCl(H 2 O) 5 ] + + H 2 O [Cu(H 2 O) 6 2+ ][Cl] [CuCl(H 2 O) 5 + ] But you know K eq = 0.28 ≠ 0.8: what does this mean? You investigate by calculation: 1.25 [CuCl(H 2 O) 5 + ]= 0.8 M, [Cu(H 2 O) 6 2+ ]= 0.8 M, [Cl] = 0.8M for this reaction: 0.8 M x 0.8 M 0.8 M It’s not at equilibrium!!

13. So under these concentration conditions: [Cu(H 2 O) 6 ] 2+ + Cl- [CuCl(H 2 O) 5 ] + + H 2 O [Cu(H 2 O) 6 2+ ][Cl] [CuCl(H 2 O) 5 + ] How will reaction system species behave? 1.25 > 0. 28 = K eq [CuCl(H 2 O) 5 + ]= 0.8 M, [Cu(H 2 O) 6 2+ ]= 0.8 M, [Cl] = 0.8M This tells you one definite thing: there’s too much in numerator, or, there’s too much product [CuCl(H 2 O) 5 + ] decreases, [Cu(H 2 O) 6 2+ ] increases, [Cl] increases

14. This is Q!!! Ratio of Concentrations under Non-Equilibrium conditions aA + bBcC + dD Q [reagents] [products] Q [A] a [B] b [C] c [D] d Q: the Reaction Quotient

15. The reaction quotient Q can be determined for any set of concentrations Possible outcomes [A] a [B] b 1. Q [C] c [D] d K eq 2. Q [A] a [B] b [C] c [D] d > K eq 3. Q [A] a [B] b [C] c [D] d < K eq