1. © DGMcC Magee Lecture

2. © DGMcC Cotton Wool Soaked in Conc. Ammonia Cotton Wool Soaked in Conc. Hydrochloric Acid NH 3(g) HCl (g) Ring of “White Smoke” NH 3(g) + HCl (g)  NH 4 Cl (s) Reversible Reactions

3. © DGMcC Moist pH Paper Heat Strongly Solid Ammonium chloride NH 4 Cl (s)  NH 3(g) + HCl (g) Reversible Reactions

4. © DGMcC NH 4 Cl (s)  NH 3(g) + HCl (g) NH 3(g) + HCl (g)  NH 4 Cl (s) Reversible Reactions NH 3(g) + HCl (g) NH 4 Cl (s)

5. © DGMcC Heat Strongly Copper sulphate crystals Hand STEA M

6. © DGMcC Reversible Reactions CuSO 4.5H 2 O  CuSO 4 + 5H 2 O CuSO 4.5H 2 O CuSO 4 + 5H 2 O CuSO 4 + 5H 2 O  CuSO 4.5H 2 O

7. © DGMcC Reversible reactions are not uncommon A+B ⇄ C+D On mixing A & B there will be no C & D ∴ the rate of the forward is high and the reverse reaction will be zero. As the reaction proceeds there is less A & B ∴ the rate of the forward reaction decreases. As C & D is formed in increasing amounts what will happen to the rate of the reverse reaction ? It will increase. Eventually what will happen to the rates? They will become equal.

8. © DGMcC When this happens we say the system is in chemical equilibrium. This is a dynamic equilibrium. The reaction starts then seems to stop. The forward and reverse reactions are both proceeding at the same rate. At equilibrium what happens to the amount of A? For every molecule of A reacting there is one formed ∴ the amount stays constant. This is true of the other reactants and products also. A+B ⇄ C+D

9. © DGMcC A+B ⇄ C+D The concentration of each substance at equilibrium will therefore remain constant We can write an equilibrium constant for this reaction in terms of the concentrations of the reactants and products K c =[C] x [D] [A] x [B] where [ ] = equilibrium concentration in moles /dm 3 Remember it is always [products] [reactants]

10. © DGMcC A+2B ⇄ 3C+4D K c =[C] 3 x [D] 4 [A] x [B] 2 where [ ] = equilibrium concentration in moles /dm 3 K p =P 3 C x P 4 D P A x P 2 B where P is the equilibrium partial pressure of a gas.

11. © DGMcC If K c or K p are small the amount of products will be small and reactants large, ∴ we say the equilibrium position is over to the left hand side (LHS). (The partial pressure of a gas in a mixture is the pressure that that gas would exert if it alone occupied the total volume of the mixture. The sum of the partial pressures of the gases in a mixture is equal to the total pressure of the mixture.) If K c or K p are large the amount of products will be large and reactants small, ∴ we say the equilibrium position is over to the right hand side (RHS). The equilibrium constants K c & K p are numerical constants for a given reaction at a given temperature.

12. © DGMcC N.B. Only dissolved substances appear in K c no gases or solids. Likewise only gases appear in K p no solids or solutions. State symbols are very important. e.g. 1.CaCO 3(s) ⇄ CaO (s) +CO 2(g) K p = e.g. 2.Fe 2+ (aq) + Ag + (aq) ⇄ Fe 3+ (aq) + Ag (s) Kc=Kc= (s), (l), (g), (aq). P CO2 [Fe 3+ (aq) ] [Fe 2+ (aq) ] x [Ag + (aq) ]

13. © DGMcC Predict how the above equilibrium position be changed by removing D adding C removing B How would the equilibrium constant be changed by removing D adding C removingB Le Chatelier’s Principle: When a change is forced upon a system in chemical equilibrium the position of the equilibrium moves so that the change is opposed. Move to RHS No change Only Temperature changes K A +B ⇄ C+D Move to LHS

14. © DGMcC Pressure is measured traditionally in atmospheres or more scientifically in N/m 2 or Pascals (Pa) 1atm = 1.01x 10 5 Pa The pressure of a gas depends only on the number of gas particles striking the walls of the container providing they are at the same temperature and in occupy the same volume. It does not depend on the nature or composition of the gas. e.g. equal amounts of nitrogen and carbon dioxide at the same temperature and in the same volume will exert the same pressure

15. © DGMcC Changes in Pressure A (g) +B (g) ⇄ C (g) 2 gas molecules 1 gas molecule increasing P will move equilibrium to RHS A (g) +B (g) ⇄ 2C (g) 2 gas molecules increasing P will have no effect A (g) +B (g) ⇄ 2C (g) + D (g) 2 gas molecules 3 gas molecules increasing P will move equilibrium to LHS

16. © DGMcC During chemical reaction heat is either given out by the reaction or heat is required by the reaction. During an exothermic reaction heat is given out During an endothermic reaction heat is taken in. Changes in Temperature

17. © DGMcC K c  1/T heat + A + B ⇄ C + D K c  T ExothermicHeat can be considered a product. Changes in Temperature (increasing T) EndothermicHeat can be considered a reactant. Equilibrium moves to LHS Equilibrium moves to RHS A + B ⇄ C + D + heat ∴ K gets smaller ∴ K gets larger

18. © DGMcC Energy Reaction Coordinate EAEA ΔH -ive Energy Level Diagrams Reactants Products This is an EXOTHERMIC reaction

19. © DGMcC Energy Reaction Coordinate EAEA ΔH +ive Energy Level Diagrams Reactants Products This is an ENDOTHERMIC reaction

20. © DGMcC The Effect of a Catalyst on K C or K P Reaction Coordinate E A cat Energy E A uncat ΔHΔH Reactants Products By providing an alternate, lower energy pathway for the reaction the catalyst effectively lowers the activation energy. Clearly the alternate pathway exists for both the forward and reverse reactions. Both are speeded up to the same extent ∴ the equilibrium position is not disturbed.

21. © DGMcC 1. Concentration (solutions only) Reaction rates vary widely, from instantaneous to very slow, from a bomb exploding to concrete setting. The factors that may influence a rate are; 2. Temperature (all) 3. Catalyst (all) 4. Particle size (solids only) 5. Pressure (gases only) 6. Light intensity (photochemical reactions only) Reaction Rates

22. © DGMcC Collision Theory All chemicals are composed of particles. For the chemicals to react the particles must collide with sufficient energy. If the energy in the collision is too little the particles will just bounce off without reacting. The minimum amount of energy required is called the activation energy (E A ). The energy in the collision will depend on the kinetic energy of the colliding particles (how fast they are moving) and their direction/orientation (steric factors). The kinetic energy of the particles in turn, depends on the temperature of the mixture.

23. © DGMcC Changing Concentration By increasing [ ] the number of particles in the given volume will increase ∴ the total number of collisions/sec will increase ∴ the number of effective collisions/sec will increase ∴ the reaction rate will increase.

24. © DGMcC Changing Temperature By increasing the temperature the total kinetic energy of the mixture is increased ∴ The individual particles, on average, will have a higher kinetic energy ∴ More collisions/sec will be effective, as more will exceed the activation energy ∴ The rate increases. Less importantly, there are more collisions as the particles are moving faster.

25. © DGMcC Catalysts A catalyst provides an alternative pathway of lower activation energy. E A cat Energy E A uncat ΔHΔH Reactants Products

26. © DGMcC Order of Reaction We know that rates generally depend on the concentration of the reactants but mathematically what is the relationship? Is rate proportional to [ ], [ ] 2,[ ] 3, [ ] ½, [ ] 0,[ ] -1, [ ] -2, [ ] -3 etc. How can we tell?

27. © DGMcC For the reaction A+2B ⇄ C+D We know Order of Reaction rate α [A] x and rate α [B] y x is called the order of reaction with respect to A y is called the order of reaction with respect to B they are small whole numbers (1,2,3,) or 0. x+y is called the order of reaction. 0 = zero order, 1 = first order, 2 = second order etc. rate  [A] x [B] y or rate = k[A] x [B] y where k = rate constant

28. © DGMcC N.B. ! For the equilibrium constant you use the stoichiometry of the equation but for rate equations you can’t. Orders can only be determined experimentally. for A+2B ⇄ 3C+4D rate = k[A][B] 2   K c =[C] 3 x [D] 4 [A] x [B] 2

29. © DGMcC The End