1. Chemistry 2402 - Thermodynamics Lecture 13 : Non Ideal Solutions and Activity Lecture 14 : Chemical Equilibria Lecture 15 : Kinetic Coefficients & the Transition State

2. Equilibrium when Conversion of Chemical Species is Allowed – i.e. Chemistry Consider the generic chemical equilibrium aA + bB  cC + dD The Gibbs free energy is given by G = N A  A + N B  B + N C  C + N D  D and a change in the Gibbs free energy is dG = dN A  A + dN B  B + dN C  C + dN D  D

3. Chemical Equilibria (cont.) Here is where the only bit of information about the chemical reaction enters. The stoichiometry requires the following relationship between the various changes in species number - The term d  is introduced just to help with notation. With this relation we can now simplify the expression for dG, dG = (c  C + d  D - a  A - b  B )d  At equilibrium dG = 0 and we have the condition c  C + d  D = a  A + b  B

4. Chemical Equilibrium in the Ideal Gas Mixture We can now substitute in our expression for the chemical potential for the ideal gas mixture,  i =  i o + k B Tln(P i /atm) which, on rearranging, gives where, N is Avogadro's number and Note that K P is independent of pressure and dependent on the ideal gas assumption.

5. Temperature Dependence of K P van't Hoff's equation (differential form) If we can assume that  H o is independent of temperature then we can integrate this equation to give van't Hoff's equation (integrated form) This result indicates that for exothermic reactions increasing the temperature, decreases K P and vice versa for endothermic reactions.

6. Flash Quiz! Using the example of an exothermic reaction, explain how van’t Hoff’s equation is consistent with the predictions of Le Châtelier’s principle.

7. Answer Using the example of an exothermic reaction, explain how van’t Hoff’s equation is consistent with the predictions of Le Châtelier’s principle. Exothermic →  H o < 0 Therefore K p goes down with increased temperature ↑T → reaction to shift to cool the system (i.e. in the endothermic direction) For an exothermic reaction, this shifts toward the reactants, so K p goes down If a change is applied to a chemical system at equilibrium; the equilibrium will shift in order to partially counteract the imposed change.

8. Chemical Equilibria In Solution Starting with the same reaction aA + bB  cC + dD we arrive at the same equilibrium condition c  C + d  D = a  A + b  B Substituting in the expression for the chemical potential of the non-ideal solution  i (sol) =  i ө (sol) + k B Tln(a i (sol)) where the reference state is a 1 molar solution of the species i in the solvent.

9. Chemical Equilibria in Solutions (cont.) This results in the following relation ΔG ө = - RTlnK a where  G ө = N(cμ C ө + dμ D ө - aμ A ө -bμ B ө ) and or Note that the expression for the equilibrium constant in terms of concentrations is only valid for the ideal solution.

10. Example: Solubility of Sparingly Soluble Salts Consider the case of a sparingly soluble salt like silver chloride AgCl(s) ↔ Ag + (aq) + Cl - (aq) At equilibrium (i.e. a saturated solution) μ AgCl (s) = μ Ag+ ө +k B Tln a(Ag + ) + μ Cl- ө + k B Tln a(Cl - ) Or a(Ag + ) a(Cl - ) = K s the solubility product

11. Sparingly Soluble Salts (cont.) Writing the solubility product in terms of activity coefficients gives K s = [γ(Ag + ) γ(Cl - )] x(Ag + ) x(Cl - ) Since the concentration of dissolved salt is low we typically find the activity coefficients are close to 1. If, however, we add extra salt with no ion in common e.g. KNO 3 then we find that the activity coefficients are less than 1. Since K s is a constant, this means the concentration of ions must increase – i.e. the solubility increases due to the added salt.

12. Sample exam questions from previous years At room temperature (25 °C), calcium carbonate has an aqueous solubility product of K sp = 4.47×10 ‑ 9 M 2. –Calculate the calcium concentration in a saturated solution of calcium carbonate in pure water at room temperature. State any assumptions or approximations you make. –Sea water has a high concentration of aqueous NaCl. This reduces the activity ratio of calcium ions to 0.405, and carbonate ions to 0.370. Calculate the calcium concentration in a saturated solution of calcium carbonate in sea water at room temperature. Using the example of an exothermic reaction, explain how van’t Hoff’s equation is consistent with the predictions of Le Châtelier’s principle. The equilibrium constant for the gas phase reaction H 2 + I 2  2HI is 45.6 at 764 K and 60.8 at 667 K. Assuming that the heat of reaction is constant over this temperature range, calculate the enthalpy change which accompanies one mole of the forward reaction.

13. Summary You should now Understand the connection between the standard free energy of a reaction, and its equilibrium constant Calculate one from the other Understand the conditions under which the integrated form of Van’t Hoff’s equation is valid Apply Van’t Hoff’s equation to calculate the change in an equilibrium constant as temperature is changed Explain how “spectator ions” can affect the solubility of a salt in a non-ideal solution Next Lecture Kinetic Coefficients & the Transition State