previously in chem104 a new chapter: equilibrium writing equilibrium expressions observing...
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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:
Keq [reagents]
[products]
[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2O
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 arrowproducts are species to the right of arrow
A quick recap on equilibrium behavior:
[Cu(H2O)6]2+ + Cl- [CuCl(H2O)5]+ + H2O
You watched:
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:
[Cu(H2O)6]2+ + Cl-
Keq [reagents]
[products]
[CuCl(H2O)5]+ + H2O
Keq [Cu(H2O)62+][Cl]
[CuCl(H2O)5+]
Pure liquids and solids don’t appear in Keq expression:
You watched:
You wrote:
0.28
When [CuCl(H2O)5+]= 0.8 M, [Cu(H2O)6
2+]= 0.4 M, [Cl] =7.1M
You know:
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
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
How large must Keq be for reaction to be “complete”?
Consider these reactions:
Keq 2.0 x 108
[Ni(NH3)6]2+ + 6H2O
[Ni(en)3]2+ + 6 NH3[Ni(NH3)6]2+ + 3 “en”
Keq 7.3 x 109
How large must Keq be for reaction to be “complete”?
Consider this reaction:
Keq 2.0 x 108
[Ni(NH3)6]2+ + 6H2O
What is Keq for this reaction: ?
K1 = 2.0 x 108[Ni(NH3)6]2+ + 6H2O[Ni(en)3]2+ + 6 NH3[Ni(NH3)6]2+ + 3 “en” K2 = 7.3 x 109
[Ni(en)3]2+ + 6 H2O
[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
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 ≠ 0.8: 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 M
0.8 M
It’s not at equilibrium!!
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
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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