Thermodynamic equilibrium constant K
• K is a dimensionless quantity.
• K can be expressed in terms of fugacity for gas phase
reactions or activities for aqueous phase reactions.
• Gas phase: Fugacity is a dimensionless quantity and
equals to the numerical value of partial pressure
expressed in bar, i.e. = pj/pθ where pθ = 1 bar).
• Aqueous phase:
1. Neutral solution: the activity, a, is equal to the numerical value
of the molality, i.e. bj/bθ where bθ = 1 mol kg-1.
2. Electrolyte solution: The activity shall now be calculated as αj =
γj*bj/bθ , where the activity coefficient, γ, denotes distance from the
ideal system where there is no ion-interactions among constituents.
The activities of solids and pure
liquids are equal to 1
•
α(solid) = 1 and α(pure liquid) = 1 (!!!)
• Illustration: Express the equilibrium constant for the heterogeneous
reaction
NH4Cl(s) ↔ NH3(g) + HCl(g)
• Answer:
In term of fugacity (i.e. thermodynamic equilibrium constant):
Kp =
In term of molar fraction (this one is not a thermodynamic equilibrium
constant): Kx =
Estimate reaction compositions at equilibrium
•
Example 1: Given the standard Gibbs energy of reaction H2O(g) → H2(g) +
1/2O2(g) at 2000K is + 135.2 kJ mol-1, suppose that steam at 200k pa is passed
through a furnace tube at that temperature. Calculate the mole fraction of O2
present in the output gas stream.
•
Solution: (details will be discussed in class)
lnK = - (135.2 x 103 J mol-1)/(8.3145 JK-1mol-1 x 2000K)
= - 8.13037
K = 2.9446x10-4
K=
( PO2 / P  )1 / 2 ( PH 2 / P  )
PH 2O / P 
Ptotal = 200Kpa
assuming the mole fraction of O2 equals x
PO2 = x* Ptotal,
PH2 = 2(x*Ptotal)
PH2O = Ptotal – PO2 – PH2 = (1-3x)Ptotal
* If a student expresses the reaction as H2O(g) → H2(g) + O2(g), can we still solve this
problem?
Equilibrium in biological systems:
• Biological standard state: pH = 7.
• For a reaction: A + vH+(aq) ↔ P
Δ rG = Δ r
Gθ
+ RT
ln(
bP
v
b A bH

1
)
= ΔrGθ + RT ln([ H  ]v )
 RT ln
bp
bA
the first two terms of the above eq. form ΔrG‡
ΔrG‡ = ΔrGθ + 7vRTln10,
ΔrG‡ is defined as Standard reaction Gibbs energy for
biochemical systems
* A better practice is to transfer the above eq into ΔrG‡ = ΔrGθ - 7vRTln10,
and then recognizes v is the stoichiometric number of H+.
Example: For a particular reaction of the form A → B
+ 2H+ in
aqueous solution, it was found that ΔrGθ = 20kJ mol-1 at 28oC. Estimate
the value of ΔrG‡.
• Solution:
ΔrG‡ = ΔrGθ - 7vRTln10
here the stoichiometric number of H+ is 2, i.e. v = 2
ΔrG‡ = 20 kJ mol-1 - 7(2)(8.3145x10-3 kJ K-1mol-1)
x(273+ 28K)ln10
= 20 kJ mol-1 – 80.676 kJ mol-1
= -61 kJ mol-1
(Notably, when measured with the biological standard, the
standard Gibbs energy of reaction becomes negative.!
A transition from endergonic to exergonic process. )
Molecular Interpretation of equilibrium
Two factors affect the thermodynamic equilibrium constant: (1) Enthalpy, and (2) Entropy.
Boltzmann distribution is independent of the nature of the particle.
The response of equilibria to
reaction conditions
• Equilibria may respond to changes in
pressure, temperature, and concentrations
of reactants and products.
• The equilibrium constant is not affected by
the presence of a catalyst.
How equilibria respond to pressure
• The thermodynamic equilibrium constant K is a
function of the standard reaction Gibbs energy, ΔrGθ .
• Standard reaction Gibbs energy ΔrGθ is defined at a
single standard pressure and thus is independent of
the pressure used in a specific reaction.
• The thermodynamic equilibrium constant is therefore
independent of reaction pressure. Such a relationship
can be expressed as:
K
( )T  0
p
• Although the thermodynamics equilibrium
constant K is independent of pressure, it does
not mean that the equilibrium composition is
independent of the pressure!!!
• Example: Consider a gas phase reaction 2A(g)
↔ B(g)
assuming that the mole fraction of A equals xA at quilibrium, then
xB = 1.0 – xA,
(1.0  xA ) Ptotal / P
K
( xA Ptotal / P ) 2
(1.0  xA ) P

2
xA Ptotal
because K does not change, xA must change in response to any
variation in Ptotal!!!
Le Chatelier’s Principle
• A system at equilibrium, when
subject to a disturbance,
responds in a way that tends to
minimize the effect of the
disturbance.
• The above statement suggests
that if the total pressure of a
system is increased, the
system will shift to the direction
that will have smaller number of
molecules, i.e. smaller pressure.
•
3H2(g) + N2(g) ↔ 2NH3(g).
Example: Predict the effect of an increase in pressure on the Haber
reaction, 3H2(g) + N2(g) ↔ 2NH3(g).
• Solution:
According to Le Chatelier’s Principle, an increase in pressure will
favor the product.
prove:
K
2
p NH
3
p N 2 pH3 2


2
2
xNH
p
total
3
3
xN 2 ptotal xH3 2 ptotal

2
xNH
3
2
xN 2 xH3 2 ptotal
Kx
2
ptotal
Therefore, to keep the thermodynamic equilibrium constant K
unchanged, the equilibrium mole fractions Kx must change by a factor of
4 if the pressure ptotal is doubled.
The response of equilibria to
temperature
• According to Le Chatelier’s Principle:
Exothermic reactions: increased temperature favors the reactants.
Endothermic reactions: increased temperature favors the products.
• The van’t Hoff equation:
r H 
RT 2
(a)
d ln K
dT

(b)
d ln K
1
d( )
T
r H 
 
R
(7.23a)
(7.23b)
Derivation of the van’t Hoff equation:
 r G
ln K  
RT
•
Differentiate lnK with respect to temperature
d ln K
1 d (  r G / T )

dT
R
dT
•
Using Gibbs-Helmholtz equation (eqn 3.53 8th edition)
d (  r G / T )
r H 

dT
T2
thus
•
d ln K
dT

r H 
RT 2
Because d(1/T)/dT = -1/T2:
d ln K
1
d( )
T
r H 
 
R
d ln K
 0 , suggesting
• For an exothermic reaction, ΔrHθ < 0, thus
dT
that increasing the reaction temperature will reduce the equilibrium
constant.