1. Review: Expressions of the thermodynamic equilibrium constant K
K, (a dimensionless quantity) can be expressed in terms of fugacities for gas phase reactions or activities for aqueous phase reactions.
Gas phase: Fugacity ( a dimensionless quantity) is equal to the numerical value of partial pressure, 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.

2. The activities of solids and pure liquid 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)
Solution:
In term of fugacity (i.e. thermodynamic equilibrium constant):
Kp =
In term of molar fraction: Kx =

3. Estimate reaction compositions at equilibrium
Example 1: Given the standard Gibbs energy of reaction H2O(g) → H2(g) + 1/2O2(g) at 2000K is 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)/( JK-1mol-1 x 2000K) =
K = x10-4
K =
Ptotal = 200Kpa
assuming the mole fraction of O2 equals x
PO2 = x* Ptotal,
PH2 = 2(x*Ptotal)
PH2O = Ptotal – PO2 – PH2 = (1-3x)Ptotal

4. Equilibria in biological systems: Standard reaction Gibbs energy for biochemical systems
Biological standard state: pH = 7.
For a reaction: A + vH+(aq) ↔ P
ΔrG = ΔrGθ + RT
= ΔrGθ + RT
the first two terms of the above eq. form ΔrG‡
ΔrG‡ = ΔrGθ vRTln10,
A better practice is to transfer the above eq into ΔrG‡ = ΔrGθ vRTln10, and then recognizes v is the stoichiometric number of H+.

5. 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 vRTln10
here the stoichiometric number of H+ is 2, i.e. v = 2
ΔrG‡ = 20 kJ mol (2)(8.3145x10-3 kJ K-1mol-1)
x( K)ln10
= 20 kJ mol-1 – 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 )

6. 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.

7. 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.

8. 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 equilibrium constant is therefore independent of reaction pressure. Such a relationship can be expressed as:

9. Example: Consider a gas reaction 2A(g) ↔ B(g)
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 reaction 2A(g) ↔ B(g)
assuming that the mole fraction of A equals xA at quilibrium, then xB = 1.0 – xA,
because K does not change, xA must change in response to any variation in Ptotal!!!

10. 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).

11. 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:
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.

12. 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:
(a) (7.23a)
(b) (7.23b)

13. Derivation of the van’t Hoff equation:
Differentiate lnK with respect to temperature
Using Gibbs-Helmholtz equation (eqn th edition)
thus
Because d(1/T)/dT = -1/T2:

14. For an exothermic reaction, ΔrHθ < 0, thus , suggesting that increasing the reaction temperature will reduce the equilibrium constant.