# TODAY A new chapter: Equilibrium Writing equilibrium expressions observing LeChatelier’s Principle Calculating equilibrium constants, K TODAY A new wrinkle.

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TODAY 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..

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

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

A quick recap on equilibrium behavior: [Cu(H 2 O) 6 ] 2+ + Cl-[CuCl(H 2 O) 5 ] + + H 2 O You observed this chemical system: When you added excess Cl- : [Cu(H 2 O) 6 ] 2+ + Cl- [CuCl(H 2 O) 5 ] + + H 2 O When you added excess H 2 O: [Cu(H 2 O) 6 ] 2+ + Cl- [CuCl(H 2 O) 5 ] + + H 2 O

The series of reactions observed: 3. [Cu(H 2 O) 6 ] 2+ + Cl-[CuCl(H 2 O) 5 ] 2+ + H 2 O To calculate K eq [Cu(H 2 O) 6 2+ ][Cl] [CuCl(H 2 O) 5 2+ ] 1. CuCl 2 (H 2 O) 2 + 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- Need K eq concentrations 0.40M x 7.1 M 0.80 M 0.28, NO UNITS! Pure liquids and solids don’t appear in K eq expression:

Combining reactions equals multiplying K’s Net CuCl 2 (H 2 O) 2 + 4H 2 O[Cu(H 2 O) 6 ] 2+ + 2 Cl- K 1 eq x K 2 eq CuCl 2 (H 2 O) 2 [CuCl(H 2 O) 5 2+ [Cl] (a) CuCl 2 (H 2 O) 2 + 3H 2 O[CuCl(H 2 O) 5 ] + + Cl- (b) [CuCl(H 2 O) 5 ] + + H 2 O[Cu(H 2 O) 6 ] 2+ + Cl- K 1 eq K net eq K 2 eq [Cu(H 2 O) 6 2+ ][Cl] [CuCl(H 2 O) 5 2+ ] [Cu(H 2 O) 6 2+ ][Cl] 2 CuCl 2 (H 2 O) 2 K net eq K 1 eq x K 2 eq

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

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

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 [Ni(en) 3 ] 2+ + 6 NH 3 [Ni(NH 3 ) 6 ] 2+ + 3 “en” K eq 7.3 x 10 9

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

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 ≠ 1.25: 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!!

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

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

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

Example problems to be used with reaction: [Cu(H 2 O) 6 ] 2+ + Cl-[CuCl(H 2 O) 5 ] + + H 2 O K eq = 0.28 A. [CuCl(H 2 O) 5 + ]= 0 M, [Cu(H 2 O) 6 2+ ]= 0.4 M, [Cl] = 0.4 M B. [CuCl(H 2 O) 5 + ]= 1 M, [Cu(H 2 O) 6 2+ ]= 1 M, [Cl] = 0.5 M C. [CuCl(H 2 O) 5 + ]= 0.01 M, [Cu(H 2 O) 6 2+ ]= 0.01 M, [Cl] = 0.01 M Compare this result with earlier equimolar at 0.8M !!

Calculations, calculations, calculations, Many types: 1. Calculating K from equil. concentrations 2. Calc’g Keq from initial and changed concentrations 3. Calc’g final concentrations from initial, change and Keq 4. Calculating a new K from adding 2 reactions

The ICEbox method For these types: 2. Calc’g Keq from initial and changed concentrations 3. Calc’g final concentrations from initial, change and Keq

Lots of K’s K c - equilibrium constant in concentrations K p - equilibrium constant in partial pressures K a - equilibrium constant for H+ dissociation K b - equilibrium constant for OH- formation K sp - equilibrium constant for solubility Not to be confused with: k – rate constants K eq – a general equilibrium constant K f - formation constant for metal complexes

The K “Zoo” KcKc KaKa KbKb little k KpKp K eq KfKf K sp

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