1. Dynamic equilibrium FromAS Reversible reactions In a closed system Both forward and reverse reactions occur at equal rates. Macroscopic properties remain constant. Macroscopic? Concentration, pressure, temperature, mass, volume i.e. properties we can see or measure. So dynamic at molecular level.

2. Week 17 © Pearson Education Ltd 2009 This document may have been altered from the original Graph to show how the concentrations of N 2 O 4 and NO 2 change until equilibrium is achieved

3. Le Chatelier’s Principle ‘Which way will the balance lie?’ Le Chatelier did cry. ‘It will move to cancel out Whichever change you bring about!’ ‘When any of the conditions affecting the position of a dynamic equilibrium are changed, then the position of the equilibrium will shift to minimise that change.’

4. The Haber Process N 2(g) + 3H 2(g)  2NH 3(g) ΔH -93kJmol -1 Complete the table for the above reaction: ChangeEffect on yield of ammonia Increase pressure Increase yield Decrease pressure Decrease yield Increase temperature Decrease yield Decrease temperature Increase yield Catalyst No change

5. Week 17 © Pearson Education Ltd 2009 This document may have been altered from the original Deduce expressions for the equilibrium constant, K c, for homogeneous reactions. Determine the units for K c.

6. The Equilibrium Law In a dynamic equilibrium the concentrations of products and reactants are determined by K c (capital K). – the equilibrium constant in terms of concentration. This relationship is given by the Equilibrium Law For a reaction: aA + bB  cC + dD K c = [C] c [D] d [A] a [B] b Where the equilibrium concentrations of reactants and products are raised to the power of the stoichiometric ratios in the balanced equation for the reaction.

7. Units of K c As with rate constants the units of K must be worked out for each reaction. E.g. For the reaction: N 2(g) + O 2(g)  2NO (g) K c = [products] [reactants] = [NO] 2 [N 2 ][O 2 ] Units = (mol.dm -3 ) 2 (mol.dm -3 )(mol.dm -3 ) NO UNITS

8. Week 17 © Pearson Education Ltd 2009 This document may have been altered from the original Calculate the value of an equilibrium constant, K c, including its units. Calculate the concentration or quantities of substances present at equilibrium.

9. Calculating K c Propanone reacts with hydrogen cyanide in the following reaction: (CH 3 ) 2 C=O + HCN  CH 3 C(OH)(CN)CH 3 A mixture initially containing 0.05mol.dm -3 propanone and 0.05mol dm -3 HCN in ethanol is left to reach equilibrium at room temp. At equilibrium the concentration of product is 0.0233 moldm -3 Calculate Kc

10. Calculating K c Since the reaction is 1:1 then 1 mole of each reactant produces 1 mole of product. Since the product concentration at equilibrium is 0.0233 mol.dm -3 The concentration of each reactant must have gone down by 0.0233 mol dm -3. propanoneHCN2-hydroxy- 2methylpropanitrile Initial conc (mol.dm -3 )0.0500 0 Equilibrium conc (mol.dm -3 ) 0.05- 0.0233= 0.0267 0.05- 0.0233= 0.0267 0.0233

11. Step 3 Write the equilibrium constant for this reaction in terms of concentrations: K c = [product] [propanone][hydrogen cyanide] = 0.0233 0.0267 x 0.0267 = 32.7 dm 3 mol -1 (Units =mol.dm -3 =1) mol.dm -3 x mol.dm -3 mol.dm -3

12. Try this: Calculate the equilibrium constant for the following reaction: H 2(g) + CO 2(g)  CO (g) + H 2 O (g) The initial concentration of hydrogen is 10.00mol.dm -3 and of carbon dioxide is 90.00 mol.dm -3. At equilibrium 9.47 mol.dm -3 of carbon monoxide are formed.

13. H2H2 CO 2 COH2OH2O Initial Mol.dm -3 10.0090.0000 Equil mol.dm -3 10.00- 9.47 = 0.53 90.00 – 9.47 = 80.53 9.47 If 9.47 mol.dm -3 of carbon monoxide are formed so too must 9.47 mol.dm -3 steam. Since the reaction is 1:1 then that concentration of each reactant must have been used up

14. Calculating K c K c = [products] [reactants] = [CO][H 2 O] [H 2 ][CO 2 ] = 9.47 x 9.47 0.53 x 80.53 = 2.10 no units

15. Try this: Find K c for the reaction between ethanol and ethanoic acid. 0.10 mol of ethanol were mixed with 0.10 mol of a solution of ethanoic acid and allowed to reach equilibrium. The total volume of the system is 20.0cm 3. We find by titration that 0.033mol ethanoic acid is present once equilibrium is reached. C 2 H 5 OH (l) + CH 3 CO 2 H (l)  CH 3 CO 2 C 2 H 5(l) + H 2 O (l)

16. Tabulate the data: (The reaction is 1:1 so every mole of acid used uses 1 mole ethanol) If started with 0.10 moles acid and 0.033 remain then (0.10-0.033) have been used to make the ester. ethanolacidesterwater Initial Moles 0.10 00 Equil moles 0.033 0.01-0.033 = 0.067 0.067 Equil Mol.dm -3 0.033/0.02 =1.65 0.067/0.02 =3.35

17. K c = [products] [reactants] = [ester][water] [acid][ethanol] = (0.067/0.02)(0.067/0.02) (0.033/0.02)(0.033/0.02) = 4.1 The volume and units cancel out so K c has no unit.

18. Ester hydrolysis: Ethyl ethanoate + water  ethanoic acid + ethanol Exactly 1 mole ester was mixed with exactly 1 mole water (from dilute HCl – the catalyst) and allowed to reach equilibrium. The equilibrium mixture was analysed and found to contain 0.300 moles of ethanoic acid. Calculate the value for K c at the temperature of the reaction. Find the equilibrium amounts not given. i.e. It’s a 1:1 reaction again. If 0.300 moles of ethanoic acid formed then so too must 0.300 moles ethanol. If 0.300 moles ethanoic acid have been formed 0.300 moles ester must have reacted.  0.700 must remain as must 0.700 moles water.

19. Fill in a table: EsterWaterEthanoic acid Ethanol Initial moles 1100 Equil moles 1-0.300 = 0.700 1-0.300 =0.700 0.300 Equil conc mol.dm -3 0.700/V 0.300/V

20. Kc = [products] [reactants] = (0.300/V)(0.300/V) (0.700/V)(0.700/V) = 0.300x0.300 0.700x0.700 = 0.184 (no units – why not?) All the V terms cancel out.

21. This question refers to the gaseous equilibrium: A (g)  2B (g) 4.0 moles of A was placed in a 20.0dm 3 container and heated to 320K until equilibrium had been established. The equilibrium mixture was found to contain 1.50 moles A. Calculate K c at this temperature. So: 4-1.5 moles A used up. Each mole of A produces 2moles B  2.50 moles A produces (2x 2.5) = 5 moles B

22. AB Initial amount (moles) 40 Equilibrium amount (moles) 1.52 x (4-1.5) =5 Equilibrium conc mol.dm -3 1.5/20 =0.07505.00/20 = 0.250

23. K c = [B] 2 [A] = (0.250) 2 0.0750 =0.833mol.dm -3

24. The Value of K c Since K c = [products] [reactants] The size of K c gives information about the position of the equilibrium. A value of 1 would indicate the position of the equilibrium is bang in the middle. When K c is >1 the equilibrium lies RIGHT. When K c is <1 the equilibrium lies LEFT. If K c is very large e.g.10 10 then the reaction is regarded as gone to completion. If K c is very small e.g.10 -10 then there is effectively no reaction.

25. The Effect of Temperature on K c Consider the following: ExampleEndothermic reaction 2HI(g)  H 2(g) +I 2(g) Exothermic reaction 2SO 2(g) + O 2(g)  2SO 3(g) Temp increase Equilibrium position shifts RIGHT – more PRODUCTS Equilibrium position shifts LEFT- more REACTANTS Temp deceaseEquilibrium position shifts LEFT- more REACTANTS Equilibrium position shifts RIGHT- more PRODUCTS

26. This means that K c MUST change since the values used to calculate it have changed. The Effect of Temperature Changes on K c Endothermic reaction. ΔH positive Exothermic reaction. ΔH negative Temperature increase K c increasesK c decreases Temperature decrease K c decreasesK c increases

27. The Effect of Concentration, Pressure or Catalyst on K c K c is ONLY affected by temperature changes. This can be seen in experimental data. Since K c controls the relative concentrations of reactants and products in a dynamic equilibrium equilibria must shift to restore K c. This means that none of the above can change K c at constant temperature.