Previously in Chem104 A new chapter: Equilibrium Writing equilibrium expressions

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 ..

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 Keq • Q: the reaction quotient • Q: how to use it

A quick recap on equilibrium expressions: [products] Keq [reagents] For this reaction: [Cu(H2O)6]2++ Cl- [CuCl(H2O)5]++ H2O 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: You watched: [Cu(H2O)6]2++ Cl- [CuCl(H2O)5]++ H2O You added excess Cl- : [Cu(H2O)6]2++ Cl- [CuCl(H2O)5]++ H2O You added excess H2O: [Cu(H2O)6]2++ Cl- [CuCl(H2O)5]++ H2O

A quick recap on equilibrium expressions: [products] Keq [reagents] You watched: [Cu(H2O)6]2++ Cl- [CuCl(H2O)5]++ H2O You wrote: [CuCl(H2O)5+] Keq 0.28 [Cu(H2O)62+][Cl] When [CuCl(H2O)5+]= 0.8 M, [Cu(H2O)62+]= 0.4 M, [Cl] =7.1M You know: Pure liquids and solids don’t appear in Keq expression:

Are Your Eyes Misleading You? What is in the graduated cylinder? Visible electronic spectra mixture can appear GREEN Abs [CuCl(H2O)5+] [Cu(H2O)62+] 700 nm 400 nm wavelength

Are Your Eyes Misleading You? What is in the graduated cylinder? Recall equilibrium concentrations: [CuCl(H2O)5+] 0.8 M Keq 0.28 0.4 M x7.1M [Cu(H2O)62+][Cl] Visible electronic spectra If Keq ~ 1, product concentrations are similar to reagent concentrations Abs mixture can appear GREEN [CuCl(H2O)5+] [Cu(H2O)62+] 700 nm 400 nm wavelength

How large must Keq be for reaction to be “complete”? Consider these reactions: [Ni(H2O)6]2++ 6 NH3 [Ni(NH3)6]2++ 6H2O Keq 2.0 x 108 [Ni(NH3)6]2++ 3 “en” [Ni(en)3]2++ 6 NH3 Keq 7.3 x 109

How large must Keq be for reaction to be “complete”? Consider this reaction: [Ni(H2O)6]2++ 6 NH3 [Ni(NH3)6]2++ 6H2O Keq 2.0 x 108

What is Keq for this reaction: ? [Ni(H2O)6]2++ 3 “en” [Ni(en)3]2++ 6 H2O [Ni(H2O)6]2++ 6 NH3 [Ni(NH3)6]2++ 6H2O K1 = 2.0 x 108 [Ni(NH3)6]2++ 3 “en” [Ni(en)3]2++ 6 NH3 K2 = 7.3 x 109 K = K1 x K2 [Ni(H2O)6]2++ 3 “en” [Ni(en)3]2++ 6 H2O K = K1 x K2 = (7.3 x 109)(2.0 x 108) = 1.5 x 1018

This Keq seems huge! What is ratio of rgt. Ni to prdt. Ni species [Ni(H2O)6]2++ 3 “en” [Ni(en)3]2++ 6 H2O K = 1.5 x 1018

What happens if the concentrations are equal: [CuCl(H2O)5+]= 0.8 M, [Cu(H2O)62+]= 0.8 M, [Cl] = 0.8M for this reaction: [Cu(H2O)6]2++ Cl- [CuCl(H2O)5]++ H2O You investigate by calculation: [CuCl(H2O)5+] 0.8 M 1.25 [Cu(H2O)62+][Cl] 0.8 M x 0.8 M But you know Keq = 0.28 ≠ 0.8: what does this mean? It’s not at equilibrium!!

So under these concentration conditions: [CuCl(H2O)5+]= 0.8 M, [Cu(H2O)62+]= 0.8 M, [Cl] = 0.8M [CuCl(H2O)5+] 1.25 > 0. 28 = Keq [Cu(H2O)62+][Cl] This tells you one definite thing: there’s too much in numerator, or, there’s too much product How will reaction system species behave? [Cu(H2O)6]2++ Cl- [CuCl(H2O)5]++ H2O [CuCl(H2O)5+] decreases, [Cu(H2O)62+] increases, [Cl] increases

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

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

Previously in Chem104 A new chapter: Equilibrium Writing equilibrium expressions