Free Energy and Chemical Equilibrium
stationary states in living systems are not thermodynamic equilibrium states
nevertheless the study of equilibrium states is useful
for biochemistry; equilibrium states play a privileged
role in thermodynamics
G=
N
X
µi n i
i =1
T = const, P = const
X
dG T,P = µi d n i
i
Second Law =⇒ dG T,P ≤ 0
equilibrium: G =min,
X
µi d n i = 0
i
general reaction
*
ν1J1 + ν2J2 + · · · −
)
− νl Jl + νm Jm + · · ·
PChem BA
4.1
write in more compact and convenient form
X
νi Ji = 0
i
νi positive for products
νi negative for reactants
extent of reaction ξ
d ni
dξ ≡
νi
independent of i due to mass balance
dG T,P =
X
µi d n i =
i
νi µi d ξ
i
(
dG T,P =
X
)
X
νi µi d ξ = ∆rGd ξ
i
equilibrium: ∆rG = 0
dynamic equilibrium
PChem BA
4.2
each forward reaction step is exactly balanced by its
back reaction
equilibrium: ∆rG =
X
νi µi = 0
i
µi = µ−i◦ + RT ln a i
X £ −◦
¤
∆rG = νi µi + RT ln a i
i
=
X
νi µ−i◦ +
X
νi RT ln a i
i
i
−
◦
= ∆rG + RT
X
ln
¡
i
= ∆rG −◦ + RT ln
Ã
Y
i
reaction quotient Q ≡
νi ¢
ai
ν
!
ai i
Y
i
ν
ai i
∆rG = ∆rG −◦ + RT lnQ
−
◦
we can determine ∆rG form tables of standard Gibbs
energies of formation
PChem BA
4.3
−
◦
standard Gibbs energy of formation, ∆fG , of substance J = standard reaction Gibbs energy (per mole
of species J ) for its formation from the elements in
their reference states; temperature is arbitrary, typi◦
cally the conventional temperature (25 C) is used
−
◦
∆rG =
X
−
◦
ν∆fG (product) −
X
ν∆fG −◦(reactant)
equilibrium
∆rG = 0
equilibrium
0 = ∆rG −◦ + RT ln K
∆rG −◦ = −RT ln K
Y ν
K ≡ Q eq = a i ,ieq equilibrium constant
i
¶
∆rG −◦
K = exp −
RT
µ
−
◦
−
◦¶
∆ r H − T ∆r S
K = exp −
RT
µ
¶
µ
−
◦
−
◦¶
∆r S
∆r H
K = exp
exp −
R
RT
µ
µ ¶
Q
∆rG = −RT ln K + RT lnQ = RT ln
K
PChem BA
4.4
If Q < K , then ∆rG < 0, i.e., the forward reaction is
spontaneous (dominates); the reaction proceeds towards the products.
If Q > K , then ∆rG > 0, and the back reaction dominates, i.e., the reaction proceeds towards the reactants.
consider ideal gas reactions
for ideal gases: a i = P i /P
−
◦
µ ¶
Y P i νi
Q=
P −◦
i
example:
*
aA + bB −
)
− cC + dD
µ ¶c µ ¶d
PC PD
P −◦ P −◦
Q = µ ¶a µ ¶b
PA PB
P −◦ P −◦
PChem BA
4.5
chemical equilibrium:
! Ã eq !d
eq c
PC
PD
Ã
K =Ã
P −◦
P −◦
! Ã eq !b
eq a
PA
PB
P −◦
P −◦
activities: see set 3
ideal gas:
ai =
Pi
P −◦
pure solid or liquid:
dilute solution:
K=
Y
i
ai = 1
a solv = 1, a solute = c solute/1 M
½
−
◦¾
∆rG
ν
a i i = exp −
RT
• equilibrium does not depend on a catalyst
PChem BA
4.6
• equilibrium constant K does not depend on total
pressure
dK
= 0,
dP
T = const
however
equilibrium composition, {x i }, generally depends on P
Pi
a i = −◦ ; and P i = x i P
P
µ
¶
µ ¶
Y ν Y x i P νi Y ν Y P νi
K = ai i =
= xi i
−
◦
P
P −◦
i
i
i
i
µ ¶∆ν
X
P
K = K x −◦
; ∆ν = νi
P
i
µ ¶−∆ν
P
K x = K −◦
P
1
or K x ∝ ∆ν
P
ideal gas
K x depends on P if ∆ν 6= 0
=⇒ equilibrium composition, x i , depends on P
PChem BA
4.7
∆ν > 0: K x & as P %: increasing pressure favors reactants
∆ν < 0: K x % as P %: increasing pressure favors
products
illustration of Le Chatelier’s principle: When a system
at equilibrium is perturbed, it responds in a way that
tends to minimize the effect of the perturbation.
• equilibrium constant K does depend on T
½
¾
d
∆rG −◦
dK
=
exp −
; P = const
dT dT
RT
½
¾ µ
¶
∆rG −◦
1 d (∆rG −◦/T )
= exp −
· −
RT
R
dT
dK
K d (∆rG −◦/T )
=−
dT
R
dT
1 dK
1 d (∆rG −◦/T )
=−
K dT
R
dT
d ln K
1 d (∆rG −◦/T )
=−
dT
R
dT
1 (−∆r H −◦)
=−
(Gibbs-Helmholtz equation)
2
R
T
PChem BA
4.8
d ln K ∆r H −◦
=
dT
RT 2
van’t Hoff equation
another illustration of Le Chatelier’s principle
∆r H −◦ < 0 (exothermic reaction under standard conditions) =⇒ K & as T %: increasing temperature favors reactants
∆r H −◦ > 0 (endothermic reaction under standard conditions) =⇒ K % as T %: increasing temperature favors products
integrate van’t Hoff equation with the assumption
−
◦
that ∆r H = const:
µ ¶
µ
¶
K2
∆r H −◦ 1
1
ln
=−
−
K1
R
T2 T1
acid-base equilibria
Brønsted-Lowry classification
+
−
acid: proton donor HA −→ H + A
+
+
base: proton acceptor B + H −→ BH
PChem BA
4.9
acid-base equilibrium in water:
−
+
*
HA(aq) + H2O(l) −
)
− A (aq) + H3O (aq)
A− conjugate base
K=
a A− · a H3O+
a HA · a H2O
dilute solutions: a H2O ≈ 1
Ka =
a A− · a H3O+
a HA
acid-ionization constant or acidity constant
if the total ion concentration is very low (rule of thumb
< 10−3 M), then
c A−
a A = −◦ = [A−], etc.
c
[A−] · [H3O+]
Ka ≈
[HA]
−
pH ≡ − log a H3O+
PChem BA
4.10
+
if γH3O+ ≈ 1, then pH = − log[H3O ]
pK a ≡ − log K a
∆rG −◦ = −RT ln K
−
◦
acid: ∆rG = −RT ln K a = −RT ln 10 · log K a
∆rG −◦ = RT ln 10 · pK a
+
−
*
B(aq) + H2O(l) −
)
− BH (aq) + OH (aq)
BH+ conjugate acid
a BH+ · a OH−
K=
a B · a H2O
dilute solutions: a H2O ≈ 1
a BH+ · a OH−
Kb =
aB
base-ionization constant or basicity constant
+
*
BH+(aq) + H2O(l) −
)
− H3O (aq) + B(aq)
PChem BA
4.11
Ka =
a H3O+ · a B
a BH+
add the two equilibria
+
−
*
2 H2O(l) −
)
− H3O (aq) + OH (aq)
K w = a H3O+ · a OH−
autoprotolysis constant or self-ionization constant of
water
−
◦
−
◦
adding two equilibria =⇒ adding ∆rG ; since ∆rG ∼
ln K =⇒ multiply the K ’s
K aK b = K w
◦
at 25 C: K w = 1.008 × 10
−14
, pK w = 14.00
pK w = pH + pOH
pOH ≡ − log a OH−
+
−
pure water: [H3O ]=[OH ]:
1
pH = pK w ≈ 7.00
2
PChem BA
4.12
pH calculations
strong acid = acid that is fully ionized in solution: molar concentration of hydronium ions = nominal molar
concentration of acid
common strong acids in water: HCl, HBr, HI, HClO4,
HNO3, H2SO4 (ionizes in two distinct steps; is a strong
acid only in its first ionization step.
strong base = base that is fully ionized in solution
common strong bases in water: LiOH, NaOH, KOH,
RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2
Group 1 hydroxides: molar concentration of hydroxide ion = nominal molar concentration of the base
Group 2 hydroxides: molar concentration of hydroxide ion = twice nominal molar concentration of the
base
weak acids or bases: only partially ionized – proton
transfer equilibrium
weak acid equilibrium: [HA] ≈ nominal molar concentration since extent of ionization is small; stoichiomePChem BA
4.13
−
+
try [A ] = [H3O ]
[H3O+]2
Ka ≈
[HA]
[H3O+] ≈ (Ka[HA])1/2
1
1
pH ≈ pK a − log[HA]
2
2
buffer solutions
aqueous solution of a weak acid and its conjugate
base
Ka =
Ka ≈
a base · a H3O+
a acid
[base] · a H3O+
[acid]
K a[acid]
+
a H3O ≈
[base]
[acid]
pH ≈ pK a − log
Henderson-Hasselbalch equation
[base]
PChem BA
4.14
conventional standard state is not appropriate for normal biological conditions
conventional standard state: a i = 1 =⇒ a H3O+ = 1 =⇒
pH = 0
biological conditions: pH u 7
use biological standard state (biochemical standard
+
◦
state) symbol
set a H3O+ = 1 for pH = 7
+
◦
connection between ∆rG and ∆rG
−
◦
+
reactants + ν H3O (aq) −→ products
∆rG +◦ = ∆rG −◦ + RT lnQ
Ã
!
Y ν
−
◦
= ∆rG + RT ln
ai i
i
Ã
−
◦
= ∆rG + RT ln
Y
j
!
νj
−ν
a j · aH
,
O+
3
Ã
−
◦
= ∆rG + RT ln
j 6= H3O+, νH3O+ = −ν
!
Y
1ν j · (10−7)−ν
j
PChem BA
4.15
+
(a j = 1: all species, except H , are in their standard state)
= ∆rG −◦ + RT ln(10−7)−ν
∆rG +◦ = ∆rG −◦ + 7νRT ln 10
+
If ν H3O (aq) occur on the product side, replace ν by
−ν in this formula
PChem BA
4.16