today a new chapter: equilibrium writing equilibrium expressions
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
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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 GenerationNo, not that Q ..
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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
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A quick recap on equilibrium expressions:
Keq [reagents]
[products]
[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2OFor 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 arrowproducts are species to the right of arrow
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A quick recap on equilibrium behavior:
[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2OYou observed this chemical system:
When you added excess Cl- :[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2O
When you added excess H2O:[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2O
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The series of reactions observed:
3. [Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]2+ + H2O
To calculate Keq [Cu(H2O)62+][Cl]
[CuCl(H2O)52+]
1. CuCl2(H2O)2 + 3H2O [CuCl(H2O)5]+ + Cl-
2. [CuCl(H2O)5]+ + H2O [Cu(H2O)6]2+ + Cl-
Need Keqconcentrations 0.40M x 7.1 M
0.80 M0.28,NO UNITS!
Pure liquids and solids don’t appear in Keq expression:
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Combining reactions equals multiplying K’s
Net CuCl2(H2O)2 + 4H2O [Cu(H2O)6]2+ + 2 Cl-
K1eq x K2
eq CuCl2(H2O)2
[CuCl(H2O)52+ [Cl]
(a) CuCl2(H2O)2 + 3H2O [CuCl(H2O)5]+ + Cl-
(b) [CuCl(H2O)5]+ + H2O [Cu(H2O)6]2+ + Cl-
K1eq
Kneteq
K2eq
[Cu(H2O)62+][Cl]
[CuCl(H2O)52+
][Cu(H2O)6
2+][Cl]2
CuCl2(H2O)2 Kneteq K1
eq x K2eq
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Are Your Eyes Misleading You?What is in the graduated cylinder?
Abs
wavelength 700 nm400 nm
[Cu(H2O)62+]
[CuCl(H2O)5+]
Visible electronicspectra
mixture can appear GREEN
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Are Your Eyes Misleading You?What is in the graduated cylinder?
Keq [Cu(H2O)6
2+][Cl]
[CuCl(H2O)5+]
Recall equilibrium concentrations:
0.280.8 M
0.4 M x 7.1M
Abs
wavelength 700 nm400 nm
[Cu(H2O)62+]
[CuCl(H2O)5+]
mixture can appear GREEN
Visible electronicspectra
If Keq ~ 1, product concentrations are similar to reagent concentrations
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How large must Keq be for reaction to be “complete”?Consider this reaction:
Keq 2.0 x 108
[Ni(H2O)6]2+ + 6 NH3 [Ni(NH3)6]2+ + 6H2O
[Ni(en)3]2+ + 6 NH3[Ni(NH3)6]2+ + 3 “en”
Keq 7.3 x 109
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What is Keq for this reaction: ?
K1 = 2.0 x 108[Ni(H2O)6]2+ + 6 NH3 [Ni(NH3)6]2+ + 6H2O[Ni(en)3]2+ + 6 NH3[Ni(NH3)6]2+ + 3 “en” K2 = 7.3 x 109
[Ni(H2O)6]2+ + 3 “en” [Ni(en)3]2+ + 6 H2O
[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
K = K1 x K2
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What happens if the concentrations are equal:
[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2O
[Cu(H2O)62+][Cl]
[CuCl(H2O)5+]
But you know Keq = 0.28 ≠ 1.25: what does this mean?
You investigate by calculation:
1.25
[CuCl(H2O)5+]= 0.8 M, [Cu(H2O)6
2+]= 0.8 M, [Cl] = 0.8M
for this reaction:
0.8 M x 0.8 M0.8 M
It’s not at equilibrium!!
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So under these concentration conditions:
[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2O
[Cu(H2O)62+][Cl]
[CuCl(H2O)5+]
How will reaction system species behave?
1.25 > 0. 28 = Keq
[CuCl(H2O)5+]= 0.8 M, [Cu(H2O)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(H2O)5+] decreases, [Cu(H2O)6
2+] increases, [Cl] increases
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This is Q!!! Ratio of Concentrations under Non-Equilibrium conditions
aA + bB cC + dD
Q[reagents]
[products]
Q [A]a[B]b
[C]c[D]d
Q: the Reaction Quotient
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The reaction quotient Q can be determined for any set of concentrations
Possible outcomes
[A]a[B]b1. Q [C]c[D]d
Keq
2. Q [A]a[B]b
[C]c[D]d
> Keq
3. Q [A]a[B]b
[C]c[D]d
< Keq
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Example problems to be used with reaction:[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2O
Keq = 0.28A. [CuCl(H2O)5
+]= 0 M, [Cu(H2O)62+]= 0.4 M, [Cl] = 0.4 M
B. [CuCl(H2O)5+]= 1 M, [Cu(H2O)6
2+]= 1 M, [Cl] = 0.5 M
C. [CuCl(H2O)5+]= 0.01 M, [Cu(H2O)6
2+]= 0.01 M, [Cl] = 0.01 M
Compare this result with earlier equimolar at 0.8M !!
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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
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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
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Lots of K’s
Kc - equilibrium constant in concentrations
Kp - equilibrium constant in partial pressures
Ka - equilibrium constant for H+ dissociation
Kb - equilibrium constant for OH- formation
Ksp - equilibrium constant for solubility
Not to be confused with:k – rate constants
Keq – a general equilibrium constant
Kf - formation constant for metal complexes
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The K “Zoo”
KcKa Kblittle kKp
Keq KfKsp