1. Chapter 17 Reaction Kinetics 17-1 The Reaction Process

2. How did you meet? Can you remember the first time you ever made a friend? What had to happen before the friendship could begin? Eye Contact Mutual Friend Accidentally Bumped into each other

3. Collision Theory In order for a reaction to occur particles must collide in: 1.A specific orientation and 2.with enough energy

4. Activation Energy The amount of energy required for a reaction to occur

5. Activation Energy Activation energy - the amount of energy the particles must have when they collide to force a reaction to occur. Activation Energy Products Reactants

6. Reaction Pathways The products have less energy than the reactants. The rxn released energy (heat) = exothermic ∆H will be negative since energy has left the system

7. Reaction Pathways The products have more energy than the reactants. The rxn absorbed energy (heat) = endothermic ∆H will be positive since energy has been added to the system

8. Practice Draw and label the energy diagram for a reaction in which ΔE = 30 kJ/mol, E a = 40 kJ/mol. Place reactants at energy level zero. Indicate determined values of ΔE forward, ΔE reverse & E a ’

9. Reaction Mechanisms Step-by-step sequence of rxns in order to obtain a final product Proposed Mechanism for Ozone Depletion via Free Chlorine Atoms Created by Decomposition of CFCs Step 1) Cl + O 3 → ClO + O 2 Step 2) 2 ClO → ClOOCl Step 3) ClOOCl → ClOO + Cl Step 4) ClOO → Cl + O 2 Proposed Mechanism for Ozone Depletion via Free Chlorine Atoms Created by Decomposition of CFCs Step 1) Cl + O 3 → ClO + O 2 Step 2) 2 ClO → ClOOCl Step 3) ClOOCl → ClOO + Cl Step 4) ClOO → Cl + O 2

10. Mechanisms Intermediates overall rxn

11. Mechanisms Slow Fast Fast Fast Rate Determining Step overall rxn

12. Catalysts vs. Intermediates overall rxn Catalysts appear 1 st as a reactant and then as a product during a mechanism. Intermediates appear 1 st as a product and then as a reactant during a mechanism.

13. Chapter 17 Reaction Kinetics 17-2 Reaction Rate

14. How can we increase the rate of a reaction? 1.Increase Surface Area 2.Increase Temperature 3.Increase Concentration 4.Increase in Pressure 5.Add a Catalyst

15. Surface Area Increase the surface area allows for a greater chance for effective collision

16. Temperature An increase in temperature will cause particles to move at a higher velocity resulting in more effective collisions

17. Concentration An increase in concentration will also cause an increase in the chance that effective collisions will occur

18. Pressure Increasing the pressure of a gas system will cause more frequent collisions

19. Catalysts Adding a catalyst lowers the amount of activation energy required

20. Catalysts Reactants Catalyst

21. Slow Rate Determining Step Rate Laws Rate = k[HBr][O 2 ] An equation that relates the rxn rate and the concentration of reactants

22. Rate Laws If no mechanism is given, then… 2H 2 + 2NO  N 2 + 2H 2 O Rate = k[H 2 ] 2 [NO] 2

23. Rate Orders 0, 1 st and 2 nd order rates Order is dependent upon what will yield a straight line 0 order 2 nd order ln [reactants] [reactants] 1 st order 1/[reactants]

24. Rate Orders 1 st order: reaction rate is directly proportional to the concentration of that reactant 2 nd order: reaction rate is directly proportional to the square of that reactant 0 order: rate is not dependant on the concentration of that reactant, as long as it is present. For Individual Components:

25. Rate Orders Overall reaction orders is equal to the sum of the reactant orders. Always determined experimentally! For Overall Order:

26. Calculating for k A + 2B  C Rate = k[A][B] 2 ExperimentInitial [A]Initial [B]Rate of Formation of C 10.20 M 2.0 x 10 -4 M/min 20.20 M0.40 M8.0 x 10 -4 M/min 30.40 M 1.6 x 10 -3 M/min What is the value of k, the rate constant?

27. Calculating for k Rate = k[A][B] 2 ExperimentInitial [A]Initial [B]Rate of Formation of C 10.20 M 2.0 x 10 -4 M/min 20.20 M0.40 M8.0 x 10 -4 M/min 30.40 M 1.6 x 10 -3 M/min 2.0 x 10 -4 = k[0.20][0.20] 2 2.0 x 10 -4 = k(0.008) k = 2.50 x 10 -2 min -1 M -2

28. Practice 1. In a study of the following reaction: 2Mn 2 O 7(aq) → 4Mn (s) + 7O 2(g) When the manganese heptoxide concentration was changed from 7.5 x 10 -5 M to 1.5 x 10 -4 M, the rate increased from 1.2 x 10 -4 to 4.8 x 10 -4. Write the rate law for the reaction. 2. For the reaction: A + B → C When the initial concentration of A was doubled from 0.100 M to 0.200 M, the rate changed from 4.0 x 10 -5 to 16.0 x 10 -5. Write the rate law & determine the rate constant for this reaction. Rate = k[Mn 2 O 7 ] 2 Rate = k[A] 2 Constant = 4.0 x 10 -3 M/s

29. More Practice 3. The following reaction is first order: CH 3 NC (g) → CH 3 CN (g) The rate of this reaction is 1.3 x 10 -4 M/s when the reactant concentration is 0.040 M. Predict the rate when [CH 3 NC] = 0.025. 4. The following reaction is first order: (CH 2 ) 3(g) → CH 2 CHCH 3 (g) What change in reaction rate would you expect if the pressure of the reactant is doubled? New Rate = 8.1 x 10 -5 M/s An increase by a factor of 2

30. Even More Practice 5. The rate law for a single step reaction that forms one product, C is R = k[A][B] 2. Write the balanced reaction of A & B to form C. 6. The rate law of a reaction is found to be R = k[X] 3. By what factor does the rate increase if the concentration of X is tripled? 7. The rate of reaction, involving 2 reactants, X & Z, is found to double when the concentration of X is doubled, and to quadruple when the concentration of Z is doubled. Write the rate law for this reaction. A + 2B → C R = k[X][Z] 2 The rate will increase by a factor of 27