1. Rates of Chemical Reactions CHEMICAL KINETICS

2. The rate of a reaction is measured by looking at the change in concentration over time. RATES OF CHEMICAL REACTIONS Finding the slope will give you the reaction rate at any given time.

3. What is the average rate between 10 and 30 seconds? What is the average rate between 30 and 40 seconds?

4. RELATIVE RATES

5. What are the relative rates of reaction for the following reaction? 2 NOCl → 2 NO + Cl 2 PRACTICE

6. Page 675 Exercise 15.2 Page 712 2, 4, 6 ASSIGNMENT

7. How will increasing temperature affect the rate? Why? How will increasing concentration affect the rate? Why? AFFECTING REACTION RATES

8. Catalysts increase the reaction rate by decreasing the activation energy(energy needed to start reaction). CATALYSTS

9. A rate law is an equation that relates concentration to the rate of reaction. For the equation: N 2 O 5 → 2 NO 2 + ½ O 2 The rate law is: rate = k[N 2 O 5 ] k is the rate constant. We can calculate its value. The rate will have units of mol/L·time RATE LAWS

10. For the general equation a A + b B → x X, the rate law is rate = k[A] m [B] n m and n have to be determined from data. The data will tell you how the rate changes when concentrations of each reactant change. Order of a reaction is related to the exponents in the rate law. In the rate law, rate = k[NO] 2 [Cl 2 ], the reaction is second order for NO, first order for Cl 2, and third order overall. DETERMINING RATE LAWS

11. Determine the rate equation and the value of k. CO + NO 2 → CO 2 + NO EXAMPLE [CO], mol/L[NO 2 ], mol/LRate, mol/L· h 5.10 X 10 -4.350 X 10 -4 3.4 X 10 -8 5.10 X 10 -4.700 X 10 -4 6.8 X 10 -8 5.10 X 10 -4.175 X 10 -4 1.7 X 10 -8 1.02 X 10 -3.350 X 10 -4 6.8 X 10 -8 1.53 X 10 -3.350 X 10 -4 10.2 X 10 -8

12. Using the rate equation, find the rate of reaction when [CO] = 3.8 X10 -4 mol/L and [NO 2 ] = 0.650X10 -4 mol/L. EXAMPLE

13. Page 683 Exercise 15.3 and 15.4 Page 713 8, 10, 12, 14 ASSIGNMENT

14. CONCENTRATION AND TIME

15. Cyclopropane will rearrange to propene in a first order process. If the initial concentration is 0.050 mol/L, how many hours would elapse if the concentration is 0.010 mol/L? k = 2.42 h -1 EXAMPLE

16. Hydrogen peroxide decomposes in a first order reaction. H 2 O 2 → H 2 O + O 2, k = 1.06 X 10 -3 min -1 What fraction of hydrogen peroxide remains after 100. min? What is the concentration after 100. min if the initial concentration was 0.020 mol/L? PRACTICE

17. Page 686 Exercise 15.5 and 15.6 Page 714 18, 19 ASSIGNMENT

18. SECOND-ORDER REACTIONS

19. The decomposition of HI is a second order reaction and k = 30. L/mol· min. How much time does it take for the concentration to drop from 0.010 mol/L to 0.0050 mol/L? PRACTICE

20. ZERO-ORDER REACTIONS

21. Regardless of the order, we should be able to create a straight line if we graph the right things. If we are not sure of the order, if the graph the three possibilities, we can determine the order. DETERMINING ORDER FROM A GRAPH

22. Zero Order First Order Second Order

23. ASSIGNMENT Page 687 Exercise 15.7 Page 689 Exercise 15.8 Page 714 20, 28, 30

24. HALF-LIFE

25. Sucrose, C 12 H 22 O 11 decomposes to fructose and glucose. If k = 0.216 h -1, what is the half life? How long will it take for 87.5% of the sample to decompose? PRACTICE

26. Radon-222 gas has a half-life of 3.8 days. If there are 4.0X10 13 atoms per liter initially, how many per liter will remain after 30 days? EXAMPLE

27. Reacting molecules must collide. The molecules must have enough energy The molecules must collide in the correct way. COLLISION THEORY

28. If the concentration increases, the reaction rate will increase. If there are more particles present, more collision will occur. EFFECT OF CONCENTRATION ON RATE

29. The activation energy is the amount of energy required to start a reaction. On an energy diagram, it is the amount of energy from the reactants to the highest point. What is the activation energy for the diagram? ACTIVATION ENERGY

30. Page 692 Exercise 15.9 Page 714 22, 24 ASSIGNMENT

31. Increasing the temperature of a sample is really increasing the energy of the particles. That means more particles have the required energy to react. In addition, more collisions will occur due to the increased speed of the particles. EFFECT OF TEMPERATURE ON RATE

32. ARRHENIUS EQUATION

33. Calculate the activation energy for the decomposition of HI if k 1 = 2.15X10 -8 at 650 K and k 2 = 2.39X10 -7 at 700K.

34. Catalysts speed up reaction rates be lowering activation energy. Even though they are not part of the equation, they are sometimes included in rate laws. Reaction take place in steps, and a catalyst takes part in certain steps, but is canceled out in the end. EFFECT OF CATALYST ON RATE

36. ASSIGNMENT Page 699 Exercise 15.10 Page 716 37 and 38

37. If a reaction takes place in steps, we call each step an elementary step. Br 2 + NO → Br 2 NOStep 1 Br 2 NO + NO → 2 BrNOStep 2 Br 2 + 2 NO → 2 BrNOOverall Reaction Each step has its own activation energy and k. Any substance that is not part of the overall reaction is a reaction intermediate. What is the reaction intermediate above? ELEMENTARY STEPS

38. Unimolecular- reactants of step include one molecule Bimolecular- reactants of step include two molecules, which could be two of the same molecule or two unique molecules Termolecular- reactants include three molecules MOLECULARITY OF ELEMENTARY STEPS

39. 3 ClO - (aq) → ClO 3 - (aq) + 2Cl - (aq) The above reaction takes place in two steps: ClO - (aq) + ClO - (aq) → ClO 2 - (aq) + Cl - (aq) ClO 2 - (aq) + ClO - (aq) → ClO 3 - (aq) + Cl - (aq) What is the molecularity of each step? Write the rate equation for each step. Show that the sum of the two steps gives the net reaction. PRACTICE

40. When a reaction has multiple steps, the slow step will determine the reaction rate. For the reaction 2 NO 2 (g) + F 2 (g) → 2 FNO 2 (g), the rate law is: rate = k[NO 2 ][F 2 ] The reaction takes place in two steps: NO 2 (g) + F 2 (g) → FNO 2 (g) + F(g) NO 2 (g) + F(g) → FNO 2 (g) Which step is slow? Which is fast? RATE-DETERMINING STEP

41. Page 705 Exercise 15.11 Page 707 Exercise 15.12 Page 716 41, 42 ASSIGNMENT

42. The rate law for an overall reaction can’t include an intermediate. This happens if the second step is slow. For the reaction 2 NO(g) + O 2 (g) → 2 NO 2 (g) the steps are: – NO(g) + O 2 (g) ⇌ OONO(g)fast, equilibrium – NO(g) + OONO(g) → 2 NO 2 (g)slow The rate law for the rate-determining step is rate = k 2 [NO][OONO] The forward and reverse rates of step one are equal. k 1 [NO][O 2 ]=k -1 [OONO], so [OONO]=k 1 /k -1 [NO][O 2 ] Substituting for [OONO], rate = k 1 k 2 /k -1 [NO] 2 [O 2 ]. This is the same as you would find experimentally. MECHANISMS WITH AN EQUILIBRIUM STEP

43. Another mechanism for 2 NO(g) + O 2 (g) → 2 NO 2 (g) is: NO(g) + NO(g) ⇌ N 2 O 2 (g)fast N 2 O 2 (g) + O 2 (g) → 2 NO 2 (g)slow Show this mechanism leads to the same rate law. EXAMPLE

44. Page 710 Exercise 15.13 Page 716 44(skip b), 46, 48, 52, 54, 62, 78, 80, 83 ASSIGNMENT