1. How is an equilibrium constant determined? Consider, the experiment you carried out and note down key steps required for any reaction

2. Key questions: How is an equilibrium constant determined? How is an equilibrium constant deduced for heterogeneous equilibria? What does K c (or K p ) tell us?

3. How is an equilibrium constant determined? 1. Combining known quantities of reactants under carefully measured and controlled conditions and allowing the reaction to come to equilibrium at a fixed temperature 2. Measuring accurately the concentration of one or more of the substances in the equilibrium mixture, and from this calculating the equilibrium concentrations of the other substances in the mixture 3. Substituting these equilibrium concentrations in the equilibrium expression and calculating the value

4. Determining Kc from our experiment results The titration reaction KCNS (aq) + AgNO 3 (aq) AgCNS (s) + KNO 3 (aq) 1. Deduce moles of CNS- and hence moles of Ag+ 2. Then work out [Ag+] in 10cm 3 solution that you titrated. Is this the concentration in the equilibrium mixture? 3. Now, continue working out the other concentrations and Kc 4. Why is the reaction slow??

5. Hydrolysis of ethyl ethanoate CH 3 COOCH 2 CH 3 + H 2 O CH 3 COOH + CH 3 CH 2 OH 1. How would you determine K c of this reaction? Suggest a method. 2. It is a very very slow reaction. Suggest how you would overcome this. 3. How could you use a computer to deduce K c for a number of experiments?

6. Example  4.6 g (0.1 mol) of ethanol and 12.0 g (0.2 mol) of ethanoic acid were mixed in a flask with 20 cm 3 of 1 moldm -3 hydrochloric acid (as a catalyst) at room temperature (20 o C, 293 K). The contents of the flask were left for a week to reach equilibrium, and then titrated with 1.0 moldm -3 sodium hydroxide; 137 cm 3 were required. A second titration of 20 cm 3 (1 moldm -3 ) hydrochloric acid required 20 cm 3 of 1.0 moldm -3 sodium hydroxide.

7. Calculating K c for this experiment. Equations required  HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l)  CH 3 COOH(l) NaOH(aq)  CH 3 COONa(aq) + H 2 O(l) Calculation Method  Titration with standard sodium hydroxide gives the total number of moles of acid (both HCl and CH 3 COOH) present at equilibrium;  The second titration provides the volume of sodium hydroxide required by the HCl catalyst alone;  Subtraction of step (2)–(1) gives the volume of sodium hydroxide required to neutralise the acid (CH 3 COOH) that has been produced by hydrolysis;  The number of moles of this acid is then calculated;  From this value, the number of moles of all species present at equilibrium can be obtained and substituted into the K c expression.

8. How is an equilibrium constant deduced for heterogeneous equilibria? CaO (s) + SO 2 (g) CaSO 3 (s) K c = ___1___ [SO 2 (g)] Explain how the K c expression has been deduced. What about K p ?

9. Key questions: How is an equilibrium constant determined? How is an equilibrium constant deduced for heterogeneous equilibria? What does K c (or K p ) tell us?

10. Example AgCl (s) + Br-(aq) AgBr(s) + Cl-(aq) The equilibrium constant for this reaction is 360 at 298K. If 0.1moldm -3 Br-(aq) is added to solid AgCl, what will be the equilibrium concentrations of Br-(aq) and Cl-(aq)? Answer: [Cl-(aq] = 0.09972moldm -3, [Br-(aq)] = 0.00028moldm -3

11. Answer AgCl (s) + Br-(aq) AgBr(s) + Cl-(aq) Initial a = 0.1 0 Eqm a-x = 0.1-x x Kc = x 0.1-x 360 = x 0.1-x 360(0.1-x) = x 36 – 360x = x 36 = x + 360x 36 = x (1 + 360) 36 = x 361 x = 0.0997 = [Cl-] and [Br-] = 0.1 – 0.0997 = 0.00028

12. What does K c (or K p ) tell us? Consider the following reactions: 2H 2 (g) + O 2 (g) 2H 2 O (g) K c = 9.1 x 10 80 mol -1 dm 3 N 2 (g) + O 2 (g) 2NO (g) K c = 4.8 x 10 -31 mol -1 dm 3 Write out Kc expressions and explain what the values are showing

13. Summary Value of KcExtent of reaction Very large K c > 1x10 10 Around 1 Very small Kc < 1 x10 -10 Reaction proceeds almost to completion, leaving very small amounts of reactants at equilibrium; equilibrium lies on the RHS Concentrations of reactants and products are nearly the same at equilibrium Reaction hardly occurs, producing very small quantities of products at equilibrium; equilibrium lies on the LHS Position of equilibrium Does Kc tell us anything about rate?

14. HW  Equilibrium questions on moodle