Thermodynamic equilibrium constant K K is a dimensionless

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 NH 4 Cl(s) ↔ NH 3(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 H 2 O(g) → H 2(g) + 1/2 O 2(g) at 2000 K is + 135. 2 k. J mol-1, suppose that steam at 200 k pa is passed through a furnace tube at that temperature. Calculate the mole fraction of O 2 present in the output gas stream. • Solution: (details will be discussed in class) ln. K = - (135. 2 x 103 J mol-1)/(8. 3145 JK-1 mol-1 x 2000 K) = - 8. 13037 K = 2. 9446 x 10 -4 K= Ptotal = 200 Kpa assuming the mole fraction of O 2 equals x PO 2 = x* Ptotal, PH 2 = 2(x*Ptotal) PH 2 O = Ptotal – PO 2 – PH 2 = (1 -3 x)Ptotal * If a student expresses the reaction as H 2 O(g) → H 2(g) + O 2(g), can we still solve this problem?

Equilibrium in biological systems: • Biological standard state: p. H = 7. • For a reaction: A + v. H+(aq) ↔ P Δr. G = Δr. Gθ + RT the first two terms of the above eq. form Δr. G‡ = Δr. Gθ + 7 v. RTln 10, Δr. G‡ is defined as Standard reaction Gibbs energy for biochemical systems * A better practice is to transfer the above eq into Δr. G‡ = Δr. Gθ 7 v. RTln 10, and then recognizes v is the stoichiometric number of H+.

Example: For a particular reaction of the form A → B + 2 H+ in aqueous solution, it was found that Δr. Gθ = 20 k. J mol-1 at 28 o. C. Estimate the value of Δr. G‡. • Solution: Δr. G‡ = Δr. Gθ - 7 v. RTln 10 here the stoichiometric number of H+ is 2, i. e. v = 2 Δr. G‡ = 20 k. J mol-1 - 7(2)(8. 3145 x 10 -3 k. J K-1 mol-1) x(273+ 28 K)ln 10 = 20 k. J mol-1 – 80. 676 k. J mol-1 = -61 k. J 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 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, Δr. Gθ. • Standard reaction Gibbs energy Δr. Gθ 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:

• Although 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 2 A(g) ↔ B(g) assuming that the mole fraction of A equals x. A at quilibrium, then x. B = 1. 0 – x. A, because K does not change, x. A 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. • 3 H 2(g) + N 2(g) ↔ 2 NH 3(g).

Example: Predict the effect of an increase in pressure on the Haber reaction, 3 H 2(g) + N 2(g) ↔ 2 NH 3(g). • Solution: According to Le Chatelier’s Principle, an increase in pressure will favor the product. prove: Therefore, to keep 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: (a) (7. 23 a) (b) (7. 23 b)

Derivation of the van’t Hoff equation: • Differentiate ln. K with respect to temperature • Using Gibbs-Helmholtz equation (eqn 3. 53 8 th edition) thus • Because d(1/T)/d. T = -1/T 2:

• For an exothermic reaction, Δr. Hθ < 0, thus , suggesting that increasing the reaction temperature will reduce the equilibrium constant.