Expressions of the equilibrium constant K The equilibrium constant, K, (a dimensionless quantity) can be expressed in terms of fugacities for gas phase reactions or activities for aqueous phase reactions. Fugacity ( a dimensionless quantity) is equal to the numerical value of partial pressure, i.e. pj/pθ where pθ = 1 bar). The activity, a, is equal to the numerical value of the molality, i.e. bj/bθ where bθ = 1 mol kg-1.

For reactions occurring in electrolyte solutions Effects of the interactions of ions on the reaction process should be considered. Such a factor can be expressed with the activity coefficient, γ, which denotes distance from the ideal system where there is no ion-interaction. The activity shall now be calculated as αj = γj*bj/bθ For a reaction A + B ↔ C + D K =

The activities of solids are equal to 1 Illustration: Express the equilibrium constant for the heterogeneous reaction NH4Cl(s) ↔ NH3(g) + HCl(g) Solution: In term of fugacity (i.e. partial pressure): Kp = In term of molar fraction: 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 = Ptotal = 200Kpa assuming the mole fraction of O2 equals x PO2 = x* Ptotal, PH2 = 2(x*Ptotal) PH2O = Ptotal – PO2 – PH2 = (1-3x)Ptotal

Equilibria in biological 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 + 7vRTln10

Solution: ΔrG‡ = ΔrGθ + 7vRTln10 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 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 (Note that when measured with the biological standard, the standard reaction Gibbs energy becomes negative!)

Molecular Interpretation of equilibrium

The response of equilibria to reaction conditions Equilibria 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 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 pressure. The equilibrium constant is therefore independent of pressure:

consider the reaction 2A(g) ↔ B(g) K is independent of pressure does NOT mean that the equilibrium composition is independent of the pressure!!! consider the reaction 2A(g) ↔ B(g) assuming that the mole fraction of A equals xA at quilibrium, then xB = 1.0 – xA, K = 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.

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 = Therefore, to keep K unchanged, the equilibrium mole fractions Kx will change by a factor of 4 if doubling the pressure ptotal.

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)

Derivation of the van’t Hoff equation: Differentiate lnK with respect to temperature Using Gibbs-Helmholtz equation (eqn 3.53 8th edition) thus Because d(1/T)/dT = -1/T2:

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

Applications of the van’t Hoff equation Provided the reaction enthalpy,ΔrHθ, can be assumed to be independent of temperature, eqn. 7.23b (or 9.26b in 7th edition) illustrates that a plot of –lnK against 1/T should yield a straight line of slope ΔrHθ/R. Example: The date below show the equilibrium constant measured at different temperatures. Calculate the standard reaction enthalpy for the system. T/K 350 400 450 500 K 3.94x10-4 1.41x10-2 1.86x10-1 1.48 Solution: 1/T 2.86x10-3 2.50x10-3 2.22x10-3 2.00x10-3 -lnK 7.83 4.26 1.68 -0.39

Continued

Example: The equilibrium constant of the reaction 2SO2(g) + O2(g) ↔ 2SO3(g) is 4.0x1024 at 300K, 2.5x1010 at 500K, and 2.0x104 at 700K. Estimate the reaction enthalpy at 500K. Solution: discussion: 1. Do we need a balanced reaction equation here? 2. What can be learned about the reaction based on the information provided? 3. Will the enthalpy become different at 300K or 700K?

Calculate the value of K at different temperatures The equilibrium constant at a temperature T2 can be obtained in terms of the know equilibrium constant K1 at T1. Since the standard reaction enthalpy is also a function of temperature, when integrating the equation 9.26b from T1 to T2, we should assume that ΔrHө is constant within that interval. so ln(K2) – ln(K1) = (9.28)

Example, The Haber reaction N2(g) + 3H2(g) ↔ 2NH3(g) At 298 K, the equilibrium constant K = 6.1x105. The standard enthalpy of formation for NH3 equals -46.1 kJ mol-1. What is the equilibrium constant at 500K? Answer: First, calculate the standard reaction enthalpy, ΔrHө, ΔrHө = 2*ΔfHө(NH3) - 3* ΔfHө(H2) - ΔfHө(N2) = 2*(-46.1) – 3*0 - 1*0 = - 92.2 kJ mol-1 then ln(K2) – ln(6.1*105) = *(-92.2*1000 J mol-1) (1/500 – 1/298) ln(K2) = -1.71 K2 = 0.18

Practical Applications of the Knowledge of the temperature dependence of the equilibrium constant (i) M(s) + 1/2O2(g) → MO(s) (ii) 1/2C(s) + 1/2O2(g) → 1/2CO2(g) (iii) C(s) + 1/2O2(g) → CO(g) (iv) CO(g) + 1/2O2(g) → CO2(g)