Chemical Kinetics & Equilibrium

1. Chemical Kinetics & Equilibrium • Chapter 16

2. Collision Model • Key Idea:Molecules must collide to react. • However, only a small fraction of collisions produces a reaction. Why? • For a reaction to occur, molecules must have: • 1. sufficient energy to break old bonds. • 2. proper orientation for collision to be effective.

3. Three possible collision orientations-- a) & b) produce reactions, while c) does not.

4. Factors Affecting Rate of Reaction • Concentration: The higher the concentration of the reactants, the more likely an effective collision will occur. • Temperature: An increase in temperature increases: • 1. the energy of a collision. • 2. the number of collisions.

5. Plot showing the number of collisions with a particular energy at T1& T2, where T2 >T1 -- Boltzman Distribution.

6. Activation Energy, Ea • Activation energy for a given reaction is a constant and not temperature dependent. • Activation energy represents the minimum energy for a reaction to occur.

7. a) The change in potential energy as a function of reaction progress. Ea is the activation energy and E is the net energy change -- exothermic. b) Molecular representation of the reaction.

8. Enthalpy -- H • Enthalpy -- at constant pressure, the change in enthalpy equals the energy flow as heat. • Exothermic -- H is negative (-). • Endothermic -- H is positive (+).

9. Catalysis • Catalyst:A substance that speeds up a reaction without being consumed • Enzyme:A large molecule (usually a protein) that catalyzes biological reactions. • Homogeneous catalyst:Present in the same phase as the reacting molecules. • Heterogeneous catalyst:Present in a different phase than the reacting molecules.

10. Energy plots for a catalyzed and an uncatalyzed pathway for an endothermic reaction.

11. Effect of a catalyst on the number of reaction-producing collisions. A greater fraction of collisions are effective for the catalyzed reaction.

12. Catalysis • The breakdown of the ozone layer is illustrated by the following equation: • Cl + O3 ---> ClO + O2 • O + ClO ---> Cl + O2 • Cl + O3 + O + ClO ---> ClO + O2 + Cl + O2 • Cl • Net Reaction: O + O3 ----> 2O2 • What is the catalyst? The intermediate?

13. Chemical Equilibrium • The state where the concentrations of all reactants and products remainconstantwith time. • On themolecular level,there is frantic activity. Equilibrium is not static, but is ahighly dynamic situation.

14. Reactions That Appear to Run to Completion • 1. Formation of a precipitate. • 2. Formation of a gas. • 3. Formation of a molecular substance such as water. • These reactions appear to run to completion, but actually the equilibrium lies very far to the right.All reactions in closed vessels reach equilibrium.

15. Molecular representation of the reaction 2NO2(g) ----> N2O4(g). c) & d) represent equilibrium.

16. Figure 16.9: (a) The initial equilibrium mixture of N2, H2 and NH3. (b) Addition of N2. (c) The new equilibrium.

17. Chemical Equilibrium • 2NO2(g) <----> N2O4(g) • The forwardreaction goes to the right. • The reverse reaction goes to the left. • At equilibrium the rate of the reverse reaction equals the rate of the forward reaction.

18. Concentration profile for the Haber Process which begins with only H2(g) & N2(g).

19. The Law of Mass Action • For • jA+kBlC + mD • The law of mass action (Cato Guldberg & Peter Waage) is represented by theequilibrium expression:

20. Equilibrium Expression • K is the equilibrium constant. • [C] is the concentration expressed in mol/L. • K is temperature dependent.

21. Equilibrium Constant, K • For an exothermic reaction, if the temperature increases, K decreases. • For an endothermic reaction, if the temperature increases, K increases.

22. Writing Equilibrium Expressions • 2O3(g) <---> 3O2(g)

23. Writing Equilibrium Expressions • Write the equilibrium expression for the following: • H2(g) + F2(g) <---> 2HF(g) • N2(g) + 3H2(g) <---> 2NH3(g)

24. Equilibrium Expression • Write the equilibrium expression for the following reaction: • 4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g)

25. Table 16.1: Results of Three Experiments for the Reaction N2(g) 1 3H2(g) 2NH3(g) at 500 ºC

26. Equilibrium Position • For a given reaction at a given temperature, there is only one equilibrium constant (K), but there are an infinite number of equilibrium positions. • Where the equilibrium position lies is determined by the initial concentrations of the reactants and products. The initial concentrations do not affect the equilibrium constant.

27. Homogeneous Equilibria • Homogeneous equilibria are equilibria in which all substances are in the same state. • N2(g) + 3H2(g) <---> 2NH3(g) • H2(g) + F2(g) <---> 2HF(g)

28. Heterogeneous Equilibria • . . .are equilibria that involve more than one phase. • CaCO3(s)  CaO(s) + CO2(g) • K = [CO2] • The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. This does not apply to gases or solutions.

29. The position of the equilibrium CaCO3(s) ---> CaO(s) + CO2(g) does not depend upon the amounts of solid CaCO3 or CaO.

30. Figure 16.10: The reaction system CaCO3(s) --> CaO(s) + CO2(g)

31. Heterogeneous Equilibria • Write equilibrium expressions for the following: • 2HOH(l) <---> 2H2(g) + O2(g) • 2HOH(g) <---> 2H2(g) + O2(g) K = [H2]2[O2]

32. Heterogeneous Equilibria • Write equilibrium expressions for the following: • PCl5(s) <---> PCl3(l) + Cl2(g) • CuSO4.5 HOH(s) <---> CuSO4(s) + 5HOH(g) K = [Cl2] K = [HOH]5

33. Le Châtelier’s Principle • . . . If a system at equilibrium is subjected to a stress, the equilibrium will be displaced in such direction as to relieve the stress.

34. Le Chatelier’s Principle • If a reactant or product is added to a system at equilibrium, the system will shift away from the added component. • If a reactant or product is removed, the system will shift toward the removed component.

35. Effect of Changes in Concentration on Equilibrium • N2(g) + 3H2(g) <---> 2NH3(g) • The addition of 1.000 M N2 has the following effect: • Equilibrium Position I Equilibrium Position II • [N2] = 0.399M [N2] = 1.348 M • [H2] = 1.197 M [H2] = 1.044 M • [NH3] = 0.203 M [NH3] = 0.304 M

36. Effect of Change in Concentration on Equilibrium • Position I • Position II K = 0.0602 K = 0.0602

37. Changes in Concentration • Predict the effect of the changes listed to this equilbrium: • As4O6(s) + 6C(s) <---> As4(g) + 6CO(g) • a) addition of carbon monoxide • b) addition or removal of C(s) or As4O6(s) • c) removal of As4(g) a) left b) none c) right

38. The Effect of Container Volume on Equilibrium • If the size of a container is changed, the concentration of the gases change. • A smaller container shifts the equilibrium to the right -- N2(g) + 3H2(g) ---> 2NH3(g). Fourgaseousmolecules produce twogaseousmolecules. • A larger container shifts to the left -- twogaseousmolecules produce fourgaseousmolecules.

39. The system of N2, H2, and NH3 are initially at equilibrium. When the volume is decreased, the system shifts to the right -- toward fewer molecules.

40. The Effect of Container Volume on Equilibrium • Predict the direction of shift for the following equilibrium systems when the volume is reduced: • a) P4(s) + 6Cl2(g) <---> 4PCl3(l) • b) PCl3(g) + Cl2(g) <---> PCl5(g) • c) PCl3(g) + 3NH3(g) <---> P(NH2)3(g) + 3HCl(g) a) right b) right c) none

41. Equilibrium Constant, K • For an exothermic reaction, if the temperature increases, K decreases. • N2(g) + 3H2(g) <---> 2NH3(g) + 92 kJ • For an endothermic reaction, if the temperature increases, K increases. • CaCO3(s) + 556 kJ <---> CaO(s) + CO2(g)

42. Effect of Temperature on Equilbirum • Predict how the equilibrium will shift as the temperature is increased. • N2(g) + O2(g) <---> 2NO(g) (endothermic) • 2SO2(g) + O2(g) <---> 2SO3(g) (exothermic) shift to the right shift to the left

43. Effects of Changes on the System • 1.Concentration:The system will shiftawayfrom the added component. • 2.Temperature: K will change depending upon the temperature [treat heat as a reactant (endothermic) and as a product (exothermic)].

44. Effects of Changes on the System (continued) • 3.Pressure: • a. Addition of inert gasdoes not affectthe equilibrium position. • b.Decreasingthe volume shifts the equilibrium toward the side with fewer gaseous molecules.

45. Figure 16.12: Shifting equilibrium by changing the temperature

47. Magnitude of K • A K value much larger than 1 means that the equilibrium system contains mostly products -- equilibrium lies far to the right. • A very small K value means the system contains mostly reactants -- equilibrium lies far to the left. • The size of K and the time required to reach equilibrium are not directly related!

48. Calculating Equilibrium Concentrations • Gaseous phosphorus pentachloride decomposes to chlorine gas and gaseous phosphorus trichloride. If K = 8.96 x 10-2 , and the equilbrium concentration of PCl5 is 6.70 x 10-3 M and that of PCl3 is 0.300 M, calculate the concentration of Cl2 at equilibrium. • PCl5(g) <---> PCl3(g) + Cl2(g)

49. Calculating Equilibrium Concentrations • PCl5(g) <---> PCl3(g) + Cl2(g) • K = 8.96 x 10-2 • [PCl5] = 6.70 x 10-3 M • [PCl3] = 0.300 M • [Cl2] = ? [Cl2] = 2.00 x 10-3 M

50. Solubility • Allows us to flavor foods -- salt & sugar. • Solubility of tooth enamel in acids. • Allows use of toxic barium sulfate for intestinal x-rays.