equilibrium constants. chemical equilibria
TRANSCRIPT
Equilibrium constants.
Chemical equilibria.
Friday, September 11, 2015SBU:CHE/PHY558, Physical & Quantitative Biology
Rutgers University: Chemical Thermodynamics
Lecturer: Gábor Balázsi
Conditions for chemical equilibrium
KA B→
Chemical equilibrium means a state where concentrations do not change in time.
Notation (the arrow does not mean a unidirectional reaction):
The arrow simply means that the equilibrium constant is:B
A
nK
n=
The differential of the Gibbs free energy at constant T and p should = 0 at equilibrium:
0A A B B
dG dN dNµ µ= + =
A B A BN N const dN dN+ = ⇒ = −( ) 0
A B AdNµ µ− =
A Bµ µ=At equilibrium the chemical potentials are equal:
Chemical potential from the partition function
The chemical potential can be obtained from the partition function through the free energy, F:
,T V
F
Nµ
∂ =
∂
lnF kT Q= −
!
Nq
QN
=ln
eqF NkT
N
≈ −
lnq
kTN
µ
= −
Partition functions for chemical reactions
0 1 2
0
...j MM
kT kT kT kT kT
j
q e e e e e
ε ε ε ε ε− − − − −
=
′ = = + + + +∑
The partition function q’ is (if we include the ground state):
ln AA
A
qkT
Nµ
′= −
The chemical potentials are given by:
1 0 2 0 0 0( ) ( ) ( )
1 ...M
kT kT kT kTq e e e e q
ε ε ε ε ε ε ε− − −− − − −
′= + + + + =
ln BB
B
qkT
Nµ
′= −
A B
A B
q q
N N
′ ′= at equilibrium.
0ε
1ε
2ε
K as a function of partition functions
0 0B A
B B B kT
A A A
N q qK e
N q q
ε ε−−′
= = =′
The equilibrium constant K is given by:
0 Aε
1 Aε
2 Aε
0 Bε
1Bε
2 Bε
A B
A B
q q
N N
′ ′=
More complex equilibria
KaA bB cC+ →
Consider the following reaction at equilibrium (a, b, c indicate stoichiometries):
The differential of the Gibbs free energy at constant T and p is:
0A A B B C CdG dN dN dNµ µ µ= + + =
AdN a dξ= −
The particle numbers are related through a “progress variable” or “reaction coordinate” :
BdN bdξ= −
CdN c dξ=
ξ
( ) 0A B C
a b c dµ µ µ ξ− − + =
A B Ca b cµ µ µ+ =
ca b
A B C
A B C
q q q
N N N
′ ′ ′=
0 0 0( )( )
( ) ( )
C A B
c c cc a bC C C
a b a b a b
A B A B A B
N q qK e
N N q q q q
ε ε ε− − −′= = =
′ ′
A lattice model of binding equilibrium
KL R C+ →
Consider the ligand L binding to the receptor R and forming a complex C:
If the volume is V (lattice sites) and concentrations are low, the partition functions are:
Rq Vz=
Lq Vz= 0 /kT
Cq Vze
ε−=
0 /kT
C C
R L R L
N q eK
N N q q Vz
ε−
= = =
V = translational partition function; z = rotational partition function (# of orientations).
K is smaller if :
V or z is large
ε0
is small
Pressure-based equilibrium constants
/( / )
( / ) ( / )
c c cD kTC C C
a b a b a b
A B A B A B
N p V kT qK e
N N p V kT p V kT q q
−∆= = =
Sometimes, pressures are easier to measure than particle numbers.
Using the ideal gas law, pV=NkT we get:
/0
0 0
( )c c
c a b D kTC Cp a b a b
A B A B
p qK kT e
p p q q
− − −∆= = where0
V=
0
0
ln ln lnq q kT p
kT kT kTN p p
µ ′ ′
= − = − =
( )lnkT pµ µ= +○
Likewise, for the chemical potential:
µ○ is the standard-state chemical potential.
It is independent of the pressure.
Le Chatelier principleThe stable equilibrium corresponds to a free energy minimum.
Perturbations away from the stable equilibrium increase F.
KA B→ Assume that NB fluctuates up by dNB. Return to equilibrium requires:
( ) 0B AdG dµ µ ξ= − ≤
0 B Adξ µ µ≥ ⇒ ≤
0B A
dξ µ µ≤ ⇒ ≥
If NB fluctuates up then µA and µB
will oppose that change.
Temperature-dependence: van’t Hoff equation
KA B→ The equilibrium state requires:
ln( ) ln( )A B A A B B
kT p kT pµ µ µ µ= ⇒ + = +○ ○
( ) ( )ln( ) ln A B A
p
B
p h T sK
p kT kT kT
µ µ µ − − ∆ ∆ − ∆= = = − = −
○ ○ ○ ○ ○
2
ln( )p
K h
T T kT kT
µ∂ ∂ ∆ ∆= − =
∂ ∂
○ ○
ln( )
(1 / )
pK h
T k
∂ ∆= −
∂
○
van’t Hoff plot
(dissociation of water)
0h∆ >○it means bonds
are breaking.
Enthalpy change from van’t Hoff plot
2
1 2 1
( ) 1 1ln
( )
p
p
K T h
K T k T T
∆= − −
○
van’t Hoff plot
4 111500 6.66 10T K K
T
− −= → = ×
4 112257 4.43 10T K K
T
− −= → = ×
22 1
1 2 1
( )( )ln 249
( )
p
p
K TR T T kJh
T T K T mol
−∆ = =
○
ln ( ) 164p
kJRT K T
molµ ∆ = − = ○
Also, at T=1500 K:
Gibbs-Helmholtz equation
p
GG H TS H G TS G T
T
∂ = − ⇒ = + = −
∂
2 2
( / ) 1 1
p p
G T G G GG T
T T T T T T
∂ ∂ ∂ = − = − −
∂ ∂ ∂
2
( / ) ( )
p
G T H T
T T
∂ = −
∂ 2
( / ) ( )
p
F T U T
T T
∂ = −
∂
These are Gibbs-Helmholtz equations, giving the sensitivity of G/T to T.
How does the free energy G(T) depend on temperature in general?
Pressure drives equilibria towards dense states
,, T pT N
V
p N
µ ∂ ∂ = ∂ ∂
ln( )p
K v
p kT
∂ ∆= −
∂
○
B Av v v∆ = −○ ○ ○
Applying pressure can shift an equilibrium.
Higher pressure shifts equilibrium
towards state with smaller volume.
ln( ) 1p B AK
p p kT kT p
µ µ µ∂ ∂ − ∂∆= = − ∂ ∂ ∂
○ ○ ○
(Maxwell relation)
ln( )0 0
p
B A
Kv v v
p
∂∆ = − < ⇒ >
∂○ ○ ○
Pressure-dependence of K
ln( )p
K v
p kT
∂ ∆= −
∂
○
B Av v v∆ = −○ ○ ○
Case #1: Ligand replaces water
molecule. Then pressure does not
affect the equilibrium.
Case #2: Ligand fills an empty
space. Then higher pressure
increases binding.