Lecture 20
Chemical Potential Reading:
Lecture 20, today: Chapter 10, sections A and B
Lecture 21, Wednesday: Chapter 10: 10‐17 – end
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Pop Question 7 – Boltzmann Distribution
Two systems with lowest energy at 0 kBT. Energy levels separated by 3 kBT in system A, and 5 kBT in B. Calculate the partition coefficients for each system, then the probabilities of finding molecules in each energy level (use the energy levels from 0 to 15 kBT). System A: System B: U = 15kBT p=0.000000291
p=0.00000030384
U = 12kBT p=0.00000584
U = 15kBT
p=0.000117
p=0.00004509330
U = 9kBT U = 10kBT
p=0.0024
p=0.006692438
U = 5kBT
U = 6kBT
p=0.0473
p=0.993245928
U = 3kBT
U = 0kBT
p=0.9501
U = 0kBT Simply apply the Boltzmann distribution:
System A: System B: Q = e0+e‐3+e‐6+e‐9+e‐12+e‐15=1.0524
Q = e0+e‐5+e‐10+e‐15=1.0068
Relationship between Q and the overall distribution of particles?
Smaller steps between levels → larger Q → molecules spread more. Less likely to find molecules at any given level
Why limit to <15 kBT? Because the contributions to the distribution (and the Q) of higher energy levels becomes MCB65
negligible – i.e. the higher energy levels are unpopulated for all practical purposes.
3/21/16 2
F0F1 ATP synthase performs an unfavorable reaction
~50 kJ/mol stored per ATP
~15,000 kJ/mol per second
Credits: John Walker
http://www.mrc-mbu.cam.ac.uk/research/atp-synthase
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Today’s goals
• Explore the link between concentration and the free energy change G 
i   

• Chemical potential Molar free energy G˚
i
n i T ,P ,n j i
• Equilibrium constant K: G o  RT ln K
Q
• Reaction quotient Q and mass action ratio Q/K: G  RT ln
K
• Using G and K to look at biological systems:
• ATP hydrolysis • Wednesday: acid‐base equilibria, protein folding
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Chemical Potential
• The chemical potential (i) is defined as the rate of change of the free energy with respect to the number of molecules:
G 
G
i   

 G(for one molecule of i)
N i T ,P ,N j i N i
Units of energy
(e.g. J)
• Often in chemical reactions, we use moles (n):
G 
i   
 Gi
n i T ,P ,n j i
Units of energy per mole
(e.g. J mol-1)
• Molar free energy, G°i
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Chemical potential at play ‐ diffusion
[ ]in = [ ]out
spontaneous
in
dNout = -dNin
in
out
ΔG <, >, or = 0?
out
equilibrium: ΔG = ?
What happens to μin and μout at equilibrium?
G 
i   
N i T ,P ,N j i
dG   dN
Figure from The Molecules of Life (© Garland Science 2008)
dG  in dN in  out dN out  0
in dN in  out dN in  0
( in  out )dN in  0
in  out
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Direction of spontaneous change and 
• System changing towards equilibrium: dG  ( in  out )dN in  0
• For a spontaneous change, dNin and (in‐out) should have opposite signs
• If Nin decreases, dNin < 0
• (in‐out) > 0 and in > out
• Molecules move spontaneously from regions of high chemical potential to low chemical potential
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Chemical potential and concentration • In an ideal dilute solution, molecules do not influence each other and the enthalpy is independent of concentration. • Assumption commonly used in biochemistry • For an ideal dilute solution, we’ll show that the difference in chemical potential is related to the ratio of concentrations: C2
  2  1  k B T ln
C1
• Where C1 and C2 are the concentrations of molecules
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G1  H1  TS1
 G1 
A1  

N
 A1 T ,P ,N B
G2  H 2  TS2
A 2
 H1 
 S1 
 
 T


N
N
 A1 T ,P,N B
 A1 T ,P ,N B
 G2 
 

N
 A 2 T ,P ,N B
 H 2 
 S2 
 
 T


N
N
 A 2 T ,P,N B
 A 2 T ,P ,N B
Figure from The Molecules of Life (© Garland Science 2008)
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For an ideal solution,  depends on entropy
  2  1

 







H
H
S
S
  1 

  2 
 T 2 
 T 1 

N A 2 T ,P ,N B 
N A1 T ,P ,N B 
N A 2 T ,P,N B
 
N A1 T ,P,N B

 H 2 
 H1 
 

ideal solution  enthalpy changes are the same  N 
N
 A 2 T ,P,N B  A1 T ,P,N B
 S2 
 S1 
  2  1  T
 T


N A 2 T ,P ,N B
N A1 T ,P,N B
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Calculating the entropy:
2
1
• We can use the probabilistic definition of entropy, with three states: NB
p2 
(B molecules)
M
N A1
p1 
(A molecules)
M
M  (N A1  N B )
p0 
(empty gridboxes)
M
Figure from The Molecules of Life (© Garland Science 2008)
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2
S1  Mk B  pi ln pi
i0
p2 does not depend on NA1  d/dNA1 = 0
 S1 

 Mk B
p0 ln p0  p1 ln p1  p2 ln p2 



N A1
N A1 T ,P,N B
 M  (N A1  N B ) M  (N A1  N B )  N A1 N A1 
 Mk B
ln
ln




 M
M 
N A1 
M
M
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Applying simple rules for derivatives (chain rule, product rule, etc..) we get:
 S1 
 N A1  M  (N A1  N B ) 
1  T
 kB T ln  ln



M
  M  
N A1 T ,P,N B
NB (solvent) >> NA1 (solute)
 S2 
 N A 2  M  (N A 2  N B ) 
2  T
 k B T ln
 ln



M
  M  
N A1 T ,P,N B
 N  M  N   N  M  N 
B
B
  2  1  k B T ln A 2  ln
 ln A1  ln

  M   M    M   M 
 N A 2  N A1 
 k B T ln
 ln 
  M   M 
C2
C 2 
 k B T ln 
C1 
C1
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Molecular diffusion decreases chemical potential
C2
  2  1  k B T ln
C1
• If C2 > C1 then ln(C2/C1) > 0 and  is positive
• The A molecules in Region 2 have a higher chemical potential
• Makes sense – molecules will move spontaneously from Region 2 (high concentration) to Region 1 (low concentration)
Figure from The Molecules of Life (© Garland Science 2008)
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Chemical potentials
• Switching to molar units:
C2
  RT ln
C1
(multiply Boltzmann constant by Avogadro’s number  R = NAkB)
• Calculating the chemical potential relative to standard state:
C
C
o
    RT ln o    RT ln
C
1

o
• ***C/C° (and therefore C/1M) is unitless
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What is chemical potential?
• Chemical potential is proportional to the logarithm of concentration:
•
Comparing two solutions:
•
Comparing to standard state:
C2
  2  1  k B T ln
C1
C
C
o
    RT ln o    RT ln
C
1
o
• In mechanics, the direction of spontaneous change is always towards a reduction in potential energy
• Similarly, in thermodynamics, the direction of spontaneous change is always towards a reduction in Gibbs free energy
• The partial molar Gibbs free energy (G˚i) of a type of molecule (i) is its “chemical potential” (˚i )
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What are the concentrations at equilibrium?
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Defining a “reaction progress variable”, ξ
• Reaction:
aA  bB  cC  dD
• Change in free energy as the reaction progresses:
dG  A dn A  B dn B  C dnC  D dn D
• These terms are NOT independent. To account for their coupling, we define the reaction progress variable (ξ or “xi”) which is a measure of how far the reaction has progressed
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Defining a “reaction progress variable”, ξ
• E.g.
2 A  1B  1C  2D
• 0 <  < 1

Figure from The Molecules of Life (© Garland Science 2008)
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Defining a “reaction progress variable”, ξ
• Reaction:
aA  bB  cC  dD
• Change in free energy as the reaction progresses:
dG  A dn A  B dn B  C dnC  D dn D
• These terms are NOT independent. To account for their coupling, we define the reaction progress variable (ξ) which is a measure of how far the reaction has progressed
dn A  a(d)
n A  n A (0)  a
n B  n B (0)  b
nC  nC (0)  c
n D  n D (0)  d
For a small
step in the
reaction:
dn B  b(d)
dnC  c(d)
dn D  d(d)
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Reaching equilibrium
• Substituting into the equation for dG:
dn i  ( /)id
dG  A dn A  B dn B  C dnC  D dn D
  A a(d)  B b(d)  C c(d)  D d(d)
 aA  bB  cC  dD d
• At equilibrium, dG = 0 and d can be non‐zero
aA  bB  cC  dD  0
aA  bB  cC  dD
• Products of chemical potential and stoichiometric
coefficients are balanced
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Equilibrium concentrations
A
B
C
D

[A]
 A o  RT ln
1
[B]
o
 B  RT ln
1
[C]
 C o  RT ln
1
[D]
o
 D  RT ln
1
a A 
b B
cc 
d D
[A], etc, refer to the equilibrium concentrations
[A]/1M is dimensionless
aA  bB  cC  dD
 o
[A]   o
[B] 
aA  aRT ln
 bB  bRT ln 

1  
1 
 o
[C]   o
[D] 

cC  cRT ln  dD  dRT ln

1  
1 

c
d
[C]
[D]
o
cC o  dD o  aA o  bB o  RT ln

G
a
b
[A]
[B]

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Defining the equilibrium constant
• We then define the equilibrium constant, K as:
c
G o  RT lnK eq
d
[C] [D]
K eq 
a
b
[A] [B]
• Keq is measurable
K eq  e
o
G RT
Equilibrium constant provides a way to determine the concentrations, the extent of reaction, at equilibrium
• Keq is unitless
• G° is in J/mol and RT is also in J/mol
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Extent of reactions at equilibrium
• Hydrolysis of ATP:
ATP + H2O  ADP + Pi + energy
• G° = ‐28.7 kJ mol‐1 at pH 7.0 and 10 mM Mg2+
[ADP][Pi ]
K
[ATP]
Extent of reaction for reaction with a smaller G°?
K  e
o
G RT
[H2O]/[H2O]° ~ 1
 e 28 / 2.478  10 5
• [Pi] in cells is maintained at ~10‐2 M, which means at equilibrium
[ADP]
[ATP]
 10 7
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G – in a situation not at equilibrium dG  aA  bB  cC  dD d  0
dG
 aA  bB  cC  dD  G
d
Figure from The Molecules of Life
(© Garland Science 2008)
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Direction of spontaneous change from observed concentrations?
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Reaction quotient, Q, describes observed conditions
• Combining:
C
    RT ln o
C

G  aA  bB  cC  dD
• We obtain:
o
G

G  cC o  dD o  aA o  bB o

o
Q  Reaction quotient
[C]cobs[D]dobs
 RT ln
[A]aobs[B]bobs

o
G

G
 RT lnQ

• Substituting:
G  RT ln K
o
Observed,
non-equilibrium
We get:
Q
Q
G  RT ln  2.3RT log
K
K
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Q/K is the mass action ratio
• The ratio Q/K is the mass action ratio
5.8 kJ/mol
o
G  RT ln K
Q
Q
G  RT ln  2.3RT log
K
K
• The mass action ratio determines whether a reaction goes forward or backward:
Q
1
K
Q
1
K
G < 0, reaction will go forward
G > 0, reaction will go backward
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Mass action ratio – Q/K
A  B
G  RT ln
Q
Q
 2.3RT log
K
K
mol-1
Figure from The Molecules of Life (© Garland Science 2008)
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G for ATP hydrolysis in cells
ATP  ADP + Pi + energy
K = [ADP][Pi]
[ATP]
• G°
= ‐28.7 kJ mol‐1
and at equilibrium:
[ADP]
 10 7
[ATP]
• In cells, [ADP]/[ATP] = 10‐3
Q
10 3
10
G  RT ln ~ RT ln 7 ~ RT ln10
K
10
1
~ 57 kJ mol
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ATP synthesis is not spontaneous
ATP  ADP + Pi + energy
K = [ADP][Pi]
[ATP]
• G°
= ‐28.7 kJ mol‐1
and at equilibrium:
[ADP]
 10 7
[ATP]
• How to drive ATP synthesis?
• From the chemical potential concept
• Increasing the concentration of ADP and Pi could lead to ATP synthesis
• Impractical for the cell
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An example of non‐expansion work
• Using a gradient of molecules across a membrane to synthesize ATP
Unfavorable reaction:
(ADP+Pi)•F
(ATP) •F*
Favorable reaction:
B(high)
B(low)
(ADP+Pi)•F + B  (ATP) •F*•B
Figure from The Molecules of Life (© Garland Science 2008)
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F0F1 ATP synthase uses proton chemical potential to synthesize ATP
Credits: John Walker
http://www.mrc-mbu.cam.ac.uk/research/atp-synthase
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Oxidative phosphorylation in mitochondria
Figures from The Molecules of Life (© Garland Science 2008)
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F0F1 ATP synthase couples ATP synthesis to a transmembrane proton gradient
(ADP+Pi)•F + H+  (ATP) •F*•H+
K=
[(ATP) •F*• H+]
[(ADP•Pi) •F][H+]
• U = q + w, and q ~ 0 (under idealized conditions)
• U = w
• U is positive because ATP is produced
• w is positive, corresponding to work done on the system by the surroundings
• Chemical work as a consequence of transferring B molecules from high to low concentration, decreasing the free energy of B, and storing this energy into synthesized ATP
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Figure from The Molecules of Life (© Garland Science 2008)
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Some concepts to remember
• The chemical potential, , describes the rate of change of the free energy with respect to concentrations
• The equilibrium constant, Keq, provides a link between free energy and the concentrations at equilibrium
• The mass action ratio, Q/K, is related to the reaction free energy change, G, determining the driving force for the reaction
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