Chemical Kinetics and Equilibrium

1. Chemical Kinetics and Equilibrium

2. Reaction Rates • How fast or slow the reaction occurs

3. Collision Theory • 2 conditions must be satisfied for a chemical reaction to occur • Particles of reactants must collide with one another; and • Colliding particles must have sufficient energy

4. Activation Energy • Minimum amount of energy needed for a reaction to occur F> Ea Ea

5. F < Ea Ea

6. Energy Diagram for Exothermic reaction increasing Average energy energy of reactants Energy difference bet. Average energy reactants and products of products

7. Energy Diagram for Endothermic reaction increasing energy energy of the products energy of reactants energy difference

8. Factors Affecting Rate of reaction • Surface Area • Concentration of reactants • Temperature • Presence of Catalyst

9. Surface area of reactants • Consider the block of wood or wood shavings • The smaller the particle size of wood the greater the surface area exposed to oxygen thus the rate of reaction increases.

10. Concentration of Reactants • An increase in concentration of the reactants causes an increase in the rate of chemical reaction

11. Temperature • Higher temperature increases reaction rates • Ex: ripe fruits are placed in the refrigerator to slow down the ripening process.

12. Presence of Catalyst • Catalyst • Substance that speeds up chemical reaction without itself undergoing a chemical change Inhibitor --->substances that slow down reaction

13. Activation energy of uncatalized reaction • catalyzed activation energy Uncatalyzed • activation energy Note: Catalysts increases • reaction rates by • providing alternative • reaction pathways • with lower activation • energies

14. Lower activation energy • Means that a lower amount of energy is needed to surpass the energy barrier.

15. Two types of catalyst • Homogeneous  Exists in the same phase as the reactant • Heterogeneous • Exist as a separate phase from the reactant in a reaction mixture • Most are solids

16. Heterogeneous and Homogeneous Reactions • Homogeneous Reaction  Reaction involving only one phase • Heterogeneous reaction  Reaction involving reactants in two phases

17. Chemical Equibrium • It is always assumed that chemical reaction go to completion • However, only few chemical reactions proceed to completion • Reversible reaction N2O4(g) 2NO

18. Do Now: • Page 547 Focus Questions 1 and 2 for 10 pts

19. Reversible Reaction • A chemical reaction in which the products can regenerate the original reactants N2O4(g) 2NO2 colorless brown N2O4(g) -------> 2NO2(g) ( forward Reaction) 2NO2 (g)-------> N2O4(g) (backward Reaction)

20. Chemical Equilibrium • The rate of forward reaction and backward reactions are equal • Is dynamic  The reactants constantly forms the products while the products constantly forms the reactants Note: In equilibrium, concentration of reactants and products do not change

21. Ex: dynamic Equilibrium • Carbonated Drinks • CO2 + H20 ---> H2CO3 carbonic acid When the bottle is opened the pressure is released, carbon dioxide evolved from the decomposition of carbonic acid H2CO3 ------> CO2 + H2O • CO2 + H20 H2CO3

22. Equilibrium Constants(K) • Can be obtained in any equilibrium reaction • Equal to the Ratio of the equilibrium concentration of the products to the equilibrium concentration of the reactants • aA + bB ----->cC + dD • Where: • a,b,c & d -- are the coefficients A,B,C & D– are the chemical species

23. Equilibrium Constant(K) aA + bB ----->cC + dD • K = [C]c [D]d = [ products ] [A]a [B]b [reactants] --> Experimentally at equilibrium, Concentration of the products raised to a certainpower divided the concentration of the reactants also raised to a certain power is constant at( fixed temperature) • Is also called equilibrium expression

24. Note: • K = [ products] [reactants] • is the concentration of products over that of the reactants • If the K value is small, it means that the equilibrium concentration of the products is small while the reactants are large. This indicates that at equilibrium, the system consist mostly of reactants • If the K value is great , it means that the equilibrium system consist mostly of the products

25. Writing equilibrium expression • If a pure liquid or solid is involved in a reaction, its concentration is omitted in the equilibrium expression because it has constant concentration in mol/L at constant Temperature

26. Ex: • Write the equilibrium expression for the reaction below • H2(g) + F2(g) 2HF(g) • K = [ HF]2 • [H2] [F2]

27. Ex:2 • Write the equilibrium expression for the reaction below • N2(g) + 3H2(g) 2NH3(g) • K = [NH3]2 [N2][H2]3

28. Ex:3 • Write the equilibrium expression for the reaction below • CaCO3(s) CaO(s) +CO2(g) • K = [ CO2]

29. Ex:3 • Write the equilibrium expression for the reaction below • 2Hg(l) + O2(g) 2HgO(s) • K = 1 since solid and liquid is omitted • [O2]

30. Ex:3 • Write the expression of the following Reactions • 2H2S(g) + 3O2(g) 2H2O(g) + 2SO2(g) • CO2(g) + H2(g) CO(g) + H2O(l) • FeO(s) + H2(g) Fe(s) + H2O(g)

31. Exercises • Write the expression of the following Reactions • 2H2S(g) + 3O2(g) 2H2O(g) + 2SO2(g) • CO2(g) + H2(g) ---> CO(g) + H2O(l) • FeO(s) + H2(g) --->

32. Calculating Equilibrium Constant (K) • Ex: For the reaction 2NBr(g) N2(g) + 3Br2(g) The system at equilibrium at a particular temperature is analyzed and the following concentrations are found: [NBr3] = 2.07 x 10-3M [N2] = 4.11x10-2M [Br2] = 1.06x 10-3M Calculate the value of K for the reaction Solution: K = [4.11x10-2M][1.06 x 10-3M]3 [2.07x 10-3M]2 K = 1.14 x 10-5 Note: the coefficients in the balanced equation are the exponents you see in the solution

33. What is Ksp? • Ksp is the sollubility product constant • It refers to a reaction in which a solid is dissolving in water Ex: AB(s)---> A+(aq) + B-(aq)

34. How are Ksp and K different? Similar? • Solubility product constant(Ksp) is a specific type of chemical equilibrium (K ). It is used when the reactant is a solid and the products are ions. In both cases the rules for setting up the equation are the same and all terms have units of mol/L

35. Writing Solubility Product Expression When CaF2 dissolves in water it dissociates and forms ions: • CaF2(s) ------> Ca2+ + F- unbalanced CaF2(s) ------> Ca2+(aq) + 2 F-(aq) balanced In writing the Ksp expression in the above reaction, the solid is being omitted so Ksp = [Ca2+] [F-]2

36. Write the balanced equation when PbCl2 solid is dissolving in water and write the Ksp expression • PbCl2(s) ----> Pb2+(aq) + Cl-(aq) PbCl2(s) ----> Pb2+(aq) + 2Cl-(aq) balanced Ksp = [Pb2+] [ Cl-]2

37. Exercises Write the balanced equation for the reaction describing the dissolving of each of the following solids in water. Also write the Ksp expression for each solid • BaSO4(s) • Fe(OH)3(s) • Ag3PO4(s)

38. Le Chatelier’s Principle • States that if a change in conditions is imposed on a system at equilibrium, the equilibrium position will shift in the direction that tends to reduce the effects of that change -->The effect of change in concentration  The effect of change in Volume The effect of change in temperature

39. Effects of Change in Concentration on reaction Equilibrium • Consider the equation below N2 + 3H2 2NH3(g) This equation shows equilibrium because of the two arrows of the same length that is rate of forward reaction is equal to the rate of backward reaction. What will happen if you increase the concentration of N2? increasing the concentration of the reactants means more collision between them that increases the rate of forward reaction as shown below N2 + 3H2 2NH3(g) Notice that the arrow pointing to the right has a greater length which indicates forward reaction increases meaning more production of NH3

40. What happens if you increase the concentration of NH3 in the previous example? N2 + 3H2 2NH3(g) Increasing the concentration of NH3 will shift the reaction to the left. Note: If a reactant or product is added to the system, the system shifts away from the added component If a reactant or product is removed, the system shifts toward the removed component

41. Exercise • Suppose the reaction system 2SO2(g) + O2(g) 2SO3 (g) Has reached equilibrium. Predict the effect of each of the following changes on the position of the equilibrium • SO2(g) is added to the system • The SO3(g) present is liquefied and removed from the system • Some of the O2 gas is removed from the system

42. The effect of Change in Volume and pressure Recall : Boyle’s law : Pressure is inversely proportional to Volume of gases The smaller the volume, the higher the pressure between molecules Avogadro’s Law : V is directly proportional to the # of moles at constant T • Consider the equation below: suppose the gases below are mixed in a vessel at equilibrium 2 NOCl(g) 2NO(g) + Cl2(g) What will happen to the equilibrium position if we reduce the volume? • Reducing volume means increasing the pressure between the molecules of the gases, the system moves in the direction that lowers its pressure. In the equation above, the direction will shift to the left because the products has more molecules (3 molecules) than the reactant (2molecules). • In other words, the equilibrium position will shift toward the side of the reaction that involves the smaller number of gaseous molecules in the balanced equation 2 NOCl(g) 2NO(g) + Cl2(g)

43. What happens when volume is increased in the previous example? 2 NOCl(g) 2NO(g) + Cl2(g) • When the volume is increased, it lowers the pressure of the system, thus the direction will shift to the right to increase the total number of gaseous molecules present

44. Predict the shift in equilibrium of the reaction below when volume is reduced a) P4(s) + 6Cl2(g) 4PCl3(l) 6 gaseous molecules 0 gaseous molecules The direction will shift toward the right since it has 0 gaseous molecules b) PCl3(g) + Cl2(g) PCl5(g) 2 gaseous molecules 1 gaseous molecule The direction will shift to the right since it has lesser number of gaseous molecules c) PCl3(g) + 3NH3(g) P(NH2)3(g) + 3HCl(g) 4 molecules 4 molecules Since the # of gaseous molecules in both sides are equal, a change in volume will have no effect on the equilibrium position

45. Practice: • For each of the following reaction predict the direction in which the equilibrium will shift when volume in the container is increased a) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) b) FeO(s) + H2(g) Fe(s) + H2O(g)

46. The effect of change in Temperature The change in concentration and volume in the system alter the equilibrium position but not the equilibrium constant(K) The effect of temperature on equilibrium is different because the value of equilibrium constant (K) is change • If the reaction is exothermic that means it releases heat the direction will shift to the left since heat is is one of the products • If the reaction is endothermic, energy is one of the reactants so the direction will shift to the right

47. Ex: • A) C(s) + O2(g) CO2(g) + heat • B) N2(g) + O2(g) 2NO(g) – endothermic Answer: A. since heat is one of the products it is exothermic thus the direction of the shift is to the right B. since it is endothermic, heat is one of the reactants so the direction of the shift is toward the left

48. Exercise: • Predict the shift would each of the changes below have on the reaction O2(g) + 2CO(g) 2CO2(g) + heat a)Increasing [CO] b)Increasing [CO2] c) Increasing Temperature d) Decreasing volume