20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #11
page 1
Equilibrium in Solution
The chemical potential for molecules in solution is given by a formula
that is very similar to that for ideal gases:
µA (T , p , cA ) = µAo (T , p ) + RT ln cA = µAo (T , p ) + RT ln [A ]
The precise definition of the standard chemical potential µAo (T , p ) is
now more complicated; it is defined at a given pH, salt concentration,
etc…, all solution properties that need to be defined in advance. We
will not go through those and take it as a given that the standard
state is appropriately defined.
Given a standard chemical potential µAo (T , p ) , then the analysis that
we did for the ideal gas follows straight through and we find for a
solution process
νA A(g, T, p) + νB B(g, T, p) = νC C(g, T, p) + νD D(g, T, p)
that following the ideal gas analysis in our previous lecture
⎛ [C ]νC [D ]νD
∆G (ε ) = ε ⎡⎣ν C µC (T ) + ν D µD (T ) ⎤⎦ − ⎡⎣ν A µA (T ) + ν B µB (T ) ⎤⎦ + RT ln ⎜
⎜ [A ]ν A [B ]ν B
⎝
o
o
o
o
⎞
⎟
⎟
⎠
and the equilibrium constant K comes out through
o
∆Grxn
= −RT ln K ,
Where K = Qeq
K = e −∆G
o
RT
[C ] [D ] at equilibrium as before, and where the
=
ν
[A ]ν [B ]
νC
νD
A
B
concentrations Q are equilibrium concentrations.
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #11
page 2
Temperature dependence of K (or Kp)
ln K (T ) = −
∆G o
RT
⇒
d ln K
d ⎛ ∆G o ⎞ ∆G o
1 d ∆G o
=
−
=
−
dT
dT ⎜⎝ RT ⎟⎠ RT 2 RT dT
But at fixed pressure and/or solutions properties (p = 1 bar, pH
constant, etc..)
d ∆G o ⎛ ∂∆G o ⎞
=⎜
⎟
dT
⎝ ∂T ⎠1 bar,pH constant, etc...
and from fundamental equation
⎛ ∂G ⎞
dG = −SdT +Vdp ⇒ ⎜
⎟ = −S
⎝ ∂T ⎠ p
∴
o
o
1
d ln K ∆H (T ) −T ∆S (T )
=
+
∆S o (T
2
dT
RT
RT
d ln K (T ) ∆H o (T
=
dT
RT 2
⎛ ∂∆G o ⎞
o
⎜
⎟ = −∆S (T
T
∂
⎝
⎠p
⇒
)
)
)
T2
∆H o (T
)dT
Integrating:
ln K (T2 ) = ln K (T1 ) + ∫
At constant p:
∆H o (T ) = ∆H o (T1 ) + ∆C p (T −T1 )
T1
T2
ln K (T2 ) = ln K (T1 ) + ∫
T1
RT
2
∆H o (T1 ) + ∆C p (T −T1 )
RT 2
dT
Over small T ranges, ∆C p (T −T1 ) can be assumed small and ∆H o
independent of T.
⇒
∆H o ⎛ 1 1 ⎞
∆H o ⎛T2 −T1 ⎞
ln K (T2 ) ≈ ln K (T1 ) +
⎜ − ⎟ = ln K (T1 ) +
⎜
⎟
R ⎝T1 T2 ⎠
R ⎝ TT
1 2 ⎠
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
If
Lecture #11
∆H o (T ) < 0
(Exothermic)
page 3
T2 >T1 means K p (T2 ) < K p (T1 )
The equilibrium shifts toward reactants
If
∆H o (T ) > 0
(Endothermic)
T2 >T1 means K p (T2 ) > K p (T1 )
The equilibrium shifts toward products
This is Le Chatelier’s principle for Temperature
•
Example: The Haber process
½ N2(g, T, p) + 3/2 H2(g, T, p) = NH3(g, T, p)
o
∆Hrxn
(298 K ) = −46.21 kJ/mol
o
∆Grxn
(298 K ) = −16.74 kJ/mol
Kp =
pNH
3
pH3 2 pN1 2
2
=p
2
−1
XNH
3
XH3 2XN1 2
2
=e
16,740 J/mol
(8.314 J/K-mol)(298 K )
= 860
2
For p = 1 bar this is pretty good, lots of product. However, the
reaction at room T is slow (this is kinetics, not thermodynamics).
Raising T to 800 K can speed it up. But since ∆H o (T ) < 0 (exothermic),
Le Chatelier tells us that the equilibrium will shift toward the
reactants.
Indeed:
What to do?
K p ( 800 K ) = 0.007
⇒
Note above
KX = p K p
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #11
page 4
Again use Le Chatelier, but with pressure! If we increase p, Eq.
shifts toward products.
⇒
Run reaction at high T and high p
For p = 1 bar, T = 800 K, Kp = 0.007
KX =
XNH
3
XH3 2XN1 2
2
But at p = 100 bar,
•
= (1 ) K p = 0.007
2
much better!
K X = (100 ) K p = 0.7
Heterogeneous Equilibria
If a product or reactant is a solid or liquid, it will not appear in the
ratio of partial p’s for Kp or in the concentrations if the equilibrium is
in solution. However, it must be used in ∆G.
Why?
Take
νA A(s) + νB B(g) = νC C(l) + νD D(g)
The solid and liquid are not mixed – they are pure states.
∆G = ⎡⎣ν C µC ( s, pure, p ) + ν D µD ( g, mix, p ) ⎤⎦ − ⎡⎣ν A µA ( l , pure, p ) + ν B µB ( g, mix, p )⎤⎦
And for (l) or (s)
⇒
µC ( pure, p ) ≈ µ o ( pure )
∆G = ν C µCo + ν D µDo − ν A µAo − ν B µBo + RT ln
(no p-dependence)
pDνD
= ∆G o + RT lnQ
νB
pB
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
∴
Lecture #11
⎡ pDνD ⎤
νB ⎥
⎣ pB ⎦Eq .
No A or C involved.
Kp = ⎢
But we still have
and
page 5
o
∆Grxn
= ν C µCo + ν D µDo − ν A µAo − ν B µBo
ln K p = −
o
∆Grxn
RT
e.g. the decomposition of limestone
CaCO3 (s) = CaO (s) + CO2 (g)
T = 25ºC
Calculate equilibrium vapor pressure at room T and elevated T.
Data at 25ºC:
Substance
CaCO3 (s)
CaO (s)
CO2 (g)
µ o (kJ/mol)
-1128.8
-604.0
-394.36
∆Hfo (kJ/mol)
-1206.9
-635.09
-393.51
At equilibrium,
∆G = µ ( CaO, s ) + µ ( CO2 , g) − µ ( CaCO3 , s )
= µ o ( CaO, s ) + µ o ( CO2 , g) + RT ln pCO2 − µ o ( CaCO3 , s )
= ∆G o + RT ln K p
where K p = pCO2 (at eq.)
The equilibrium constant includes only the gas, but ∆G o includes
the solids too.
∆G o (kJ/mol) = -604.0 – 394.4 – (-1128.8) = 130.4 kJ/mol
∆H o (kJ/mol) = -635.1 – 393.5 – (-1206.9) = 178.3 kJ/mol
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #11
page 6
Equilibrium pressure:
ln K p = −
∆G o
RT
=−
130, 400 J/mol
= −52.50
(8.314 J/K-mol) (298.15 K )
K p = 1.43x 10 −23 bar
Nothing there at room T !
Try 1100 K:
∆H o ⎛ 1
1 ⎞
−
⎜
R ⎝ 1100 K 298 K ⎟⎠
178, 300 J/mol ⎛ 1
1 ⎞
= −52.50 −
−
⎜
⎟ = 0.17
8.314 J/K-mol ⎝ 1100 K 298 K ⎠
(1100 K ) ≈ 0.84 bar
ln pCO2 (1100 K ) ≈ ln pCO2 (298 K ) +
pCO
2
There’s probably some change in ∆Hfo over such a wide T range,
but clearly the equilibrium shifts dramatically.