9/14/16
Thermodynamics &
Statistical Mechanics
2016
aA + bB ßà cC + dD
http://chemwiki.ucdavis.edu/Physical_Chemistry/Chemical_Equilibrium/The_Equilibrium_Constant/
Calculating_An_Equilibrium_Concentration_From_An_Equilibrium_Constant/
Writing_Equilibrium_Constant_Expressions_involving_solids_and_liquids
http://boomeria.org/chemlectures/textass2/secondsemass.html
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Red + Blue ßà red-blue
∆G° =-RT ln Keq
Keq = e-∆G°/RT
http://www.chegg.com/homework-help/questions-and-answers/ba-chemical-reaction-red-blue-produces-red-blue--compounds-agaseous-state-picture-shown-re-q577493
∆G° =-RT ln Keq
Keq = RT ln [red-blue]
[red][blue]
B + R ßà B-R
At equilibrium there are
3 R-R; 5 B-B and 7 R-B
Keq is 7/(3)(5)
∆G° = -RTln(7/15) = 0.45kcal/mol
∆G°= -1.36 log(7/15)
Positive so there are more reactant than
product at equilibrium
Note: This should be B+R à2B-R (calculate ∆G° for that reaction)
∆G° =-RT ln Keq
∆G = ∆G° + RT ln [C]c[D]d
[A]a[B]b
Keq = e-∆G°/RT
- 
- 
What if there were one mole of B, R, and B-R?What would be the ∆G be?
If there were 3 red; 5 blue and 7 r-b what is ∆G?
B + R ßà B-R
At equilibrium there are
3 red; 5 blue and 7 R-B
∆G° = -RTln(7/15) = 0.45
aA + bB ßà cC + dD
Keq = RT ln [C]c[D]d
[A]a[B]b
Keq = e-∆G°/RT
Pure R or P are at high energy
(actually they are undefined
since they would be log(0) or
log(1/0)
https://wikispaces.psu.edu/pages/viewpage.action?
pageId=112526687&navigatingVersions=true
The equilibrium mixture (Keq) is
at the lowest energy.
I will use 1.36 or 1.4 for this constant
This number depends on T(since it has RT in it)
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∆G° =-RT ln Keq
∆G°=-1.36 log Keq
∆G°
Kcal/
mol
13.6
12.24
10.88
9.52
8.16
6.8
5.44
4.08
2.72
1.36
0
-1.36
-2.72
-4.08
-5.44
-6.8
-8.16
-9.52
-10.88
Keq
1E-10
0.000000001
0.00000001
0.0000001
0.000001
0.00001
0.0001
0.001
0.01
0.1
1
10
100
1000
10000
100000
1000000
10000000
100000000
∆G°=-1.36 log Keq
Every step of 1.36 kcal/mol changes the
equilibrium constant by 10 fold.
∆G°
Kcal/
mol
13.6
12.24
10.88
9.52
8.16
6.8
5.44
4.08
2.72
1.36
0
-1.36
-2.72
-4.08
-5.44
-6.8
-8.16
-9.52
-10.88
∆G°
Kcal/
mol
13.6
12.24
10.88
9.52
8.16
6.8
5.44
4.08
2.72
1.36
0
-1.36
-2.72
-4.08
-5.44
-6.8
-8.16
-9.52
-10.88
Keq
1E-10
0.000000001
0.00000001
0.0000001
0.000001
0.00001
0.0001
0.001
0.01
0.1
1
10
100
1000
10000
100000
1000000
10000000
100000000
∆G° =-RT ln Keq
∆G°=-1.36 log Keq
Every step of 1.36
kcal/mol changes
the equilibrium
constant by 10 fold.
Keq
1E-10
0.000000001
0.00000001
0.0000001
0.000001
0.00001
0.0001
0.001
0.01
0.1
1
10
100
1000
10000
100000
1000000
10000000
100000000
∆G°=-1.36 log Keq
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9/14/16
∆G°
Kcal/
mol
13.6
12.24
10.88
9.52
8.16
6.8
5.44
4.08
2.72
1.36
0
-1.36
-2.72
-4.08
-5.44
-6.8
-8.16
-9.52
-10.88
Keq
1E-10
0.000000001
0.00000001
0.0000001
0.000001
0.00001
0.0001
0.001
0.01
0.1
1
10
100
1000
10000
100000
1000000
10000000
100000000
∆G°=-1.36 log Keq
If reaction is AßàB what is B/A
If Keq = 1, 10, 100, 0.1, 0.001
What is ∆G°?
R àP
P•E
http://www.miniphysics.com/2011/11/uy1maxwell-boltzmann-distribution.html
R•E
R+E àR•EàP•EàP+E
(catalyzed reaction
R•E (reactant bound
to enzyme)
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9/14/16
What if the P/R ratio does not equal Keq?
The energy of the system has changed
∆G= ∆G° + 1.4 kcal/mol log [P]/[R]
∆G= -1.4 log Keq + 1.4 kcal/mol log [P]/[R]
(so ∆G = 0 when P/R = Keq (when the
concentrations are at equilibrium)
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