Equilibrium Revision

1. Writing equilibrium constant expressions Answers What does K tell us? Le Chatelier’s principle Question One – Le Chatelier’s Principle Answers to Question One Question Two – Le Chatelier’s Principle Answers to Question Two Question Three – Le Chatelier’s Principle Answers to Question 3 Questions – K values Answers – K values The moral Equilibrium Revision

2. start Writing equilibrium constant expressions Write the Equilibrium constant expressions for the following reactions. • N2(g) + 3H2(g)  2NH3(g) • 2SO2(g) + O2(g)  2SO3(g) • 2NO2(g)  N2O4(g) • CaCO3(s)  CO2(g) + CaO(s) • 2Ag+(aq) + Cu(s)  Cu2+(aq)+ 2Ag(s) • HCl (aq) +H2O(l)  H3O+(aq) + Cl-(aq)

3. start Answers [NH3]2 [N2] x [H2]3 1 2 3 4 5 6 [SO3]2 [O2] x [SO2]2 [N2O4] [NO2]2 [CO2] [Cu2+] [Ag+]2 [H3O+][Cl-] [HCl]

4. start What does K tell us? • If K is large, it tells us the position of equilibrium lies to the right – product favoured reaction • If K is small, it tells us the position of equilibrium lies to the left – reactant favoured reaction • If K is 1, it tells us the position of equilibrium lies in the centre – equal products and reactants

5. start Le Chatelier’s principle • If a change is imposed on a system at equilibrium, the system reacts to try and minimise the change. • An increase in concentration will produce a move in the direction to reduce the concentration • An increase in pressure (by decreasing the volume) will produce a move in the direction of the smallest number of moles of gas • An increase in temperature will produce a movement in the endothermic direction • A catalyst doesn’t produce a move – but speeds up both forward and reverse reactions

6. start Question One – Le Chatelier’s Principle In the gas phase equilibrium described by the equation below, state what would happen to the equilibrium and give reasons for your answer. N2O4(g)  2NO2(g) Kc = 4.6 x 10-3 1. Some extra N2O4 gas is injected into the equilibrium mixture. 2. Increasing the pressure of the equilibrium mixture. 3. Adding a catalyst. 4. N2O4 gas is colourless and NO2 gas is brown. Describe any colour change noticed when the pressure on the equilibrium is decreased.

7. start Answers to Question One 1. More NO2 would form/Reaction will shift to the right. Increasing the concentration of the N2O4 will increase the rate of the forward reaction and cause more products to be formed. 2. More N2O4 will be formed/Reaction will shift to the left. In this gas phase reaction there are two moles of gaseous products compared to one mole of gaseous reactants so increasing the pressure will shift the equilibrium to the side with fewer gaseous molecules. 3. The catalyst will not affect the position of the equilibrium. However it will increase the rate of both the forward and reverse reactions causing it to reach equilibrium much faster. 4. Decreasing the pressure will cause the mixture to be a darker brown.

8. start Question Two – Le Chatelier’s Principle For the reaction in Q1 [N2O4(g)  2NO2(g)] the ∆rH value is 54 kJmol-1. 1. Is this reaction exothermic or endothermic? 2. Discuss what you would observe if an equilibrium mixture of these two gases was heated.

9. start Answers to Question Two 1. Endothermic. 2. As the mixture was heated the colour would become a darker brown. Heat will favour the endothermic forward reaction producing more NO2 gas and increasing the intensity of the brown colour in the mixture.

10. start Question Three – Le Chatelier’s Principle • In the gas phase equilibrium described by the equation below, state what would happen to the equilibrium and give reasons for your answer. • N2(g) + 3H2(g)  2NH3(g) ∆rH = - 92 kJmol-1. • The reaction mixture was cooled. • The pressure was increased. • A catalyst was added. • Ammonia was removed.

11. start Answers to Question 3 • Cooling the reaction mixture will increase the equilibrium concentration of NH3 gas. This is because the exothermic forward reaction will be favoured by the removal of heat. • Increasing the pressure will increase the equilibrium concentration of NH3 gas. This is because the forward reaction involves a reduction in the number of gaseous molecules from 4 to 2. This will be favoured by an increase in pressure. • The catalyst will not affect the position of the equilibrium. However it will increase the rate of both the forward and reverse reactions causing it to reach equilibrium much faster. • Removal of ammonia will cause more ammonia to be produced. Removing ammonia decreases the rate of the reverse reaction and so the forward reaction will predominate.

12. start Questions – K values • For the equilibrium: • 2SO2(g) + O2(g)  2SO3(g) DH -ve • What would happen to the value of the equilibrium constant, K if; • The concentration of SO2 was increased • The total pressure was increased by decreasing the volume of the container • The temperature was increased • A catalyst was introduced

13. start Answers – K values • No change – all concentrations return to the position of equilibrium at this temperature • No change – changing the volume changes the concentration of all components – the ratio will be the same. • The K value would decrease as the concentration of reactants increases and the concentration of products decreases • No change – both forward and reverse reactions speed up but there is no change in K

14. start The moral • Only a change in temperature affects the value of K.