1. Kinetics Macroscopic view: How fast? Rates of reaction typically as  concentration per time From properties of materials to Reaction Chemistry Microscopic view: What path? Mechanism sequence of chemical steps, to control it 2 levels of study, for 2 reasons: Let’s try it…

2. H 2 O 2 decomposition reaction 2 H 2 O 2 (aq) --> O 2 (g) + 2 H 2 O

3. The Bombardier beetle in action

5. H 2 O 2 decomposition reaction 2 H 2 O 2 (aq) --> O 2 (g) + 2 H 2 O reaction progress affected by: o KIfaster Clno effect o Fe(3+)faster o Cu(2+)faster Also: higher [H 2 O 2 ]faster higher [KI]faster

6. How to express ‘faster’ and ‘slower’? Rate =  M /  time (for solutions) so units of reaction Rate in M/sec or M sec -1 or mol L -1 sec -1 conventions: Rate is positive – so disappearance of peroxide has negative rate 2 H 2 O 2 (aq) --> O 2 (g) + 2 H 2 O

7. How to express ‘faster’ and ‘slower’? conventions: Rate =  M /  time describes Average Rate [H 2 O 2 ] o Time, sec

8. How to express ‘faster’ and ‘slower’? conventions: Instantaneous Rate: measured over infinitely small times, a differential function: Rate =  M /  time more precise [H 2 O 2 ] o Time, sec

9. How to express ‘faster’ and ‘slower’? conventions: Rate depends on stoichiometry 2 H 2 O 2 (aq) --> O 2 (g) + 2 H 2 O More generally, for: A + B  C + D Rate = +  C]/  t = +  D]/  t = -  A]/  t = -  B]/  t In units mol / L. sec

10. [H 2 O 2 ] o Time, sec conventions: Initial Rates depend on initial concentrations

11. Experiment to obtain kinetic data to measure H 2 O 2 decomposition

12. Let’s use these conventions and look at some real data for the peroxide decomposition

13. Data: O 2 pressure as H 2 O 2 decomposes over 10 min Note: Non-linear Plot means Rate not the same at beginning and at end

14. For small time plot of data is nearly linear, so  pO 2 /  time approaches Instantaneous Rate In first 0.10 sec, the pressure goes from 102.74 to 102.91 kPA  pO 2 /  time = Rate (102.91 - 102.74) kPA / 0.10 min Rate = 1.7 kPA / min = slope

15. In first 0.10 sec, the Rate = 1.7 kPA / min After 4 min, a 0.10 sec interval shows the pressure goes from 107.49 to 107.59 kPA  pO 2 /  time = Rate (107.59 - 107.49 ) kPA / 0.10 min Rate = 1.0 kPA / min = slope For small time plot of data is nearly linear, so  pO 2 /  time approaches Instantaneous Rate

16. Data on how the rate of H 2 O 2 decomposition is affected by varying the initial [H 2 O 2 ] 2X 4.1 X Initial [H 2 O 2 ] is related to rates. What does a plot of [H 2 O 2 ] o vs Rate look like?

17. Then: Rate = k[H 2 O 2 ] o or Rate / [H 2 O 2 ] o = k units: M / s M -1 = sec -1 Average k = 8.5 x10 -4 sec -1 from a line fitting of data: Rate constant, k = 8.3 x10 -4 sec -1 2.9 x10 -4 /.35 = 8.3 x10 -4 sec -1 2.15 x10 -4 /.25 = 8.6 x10 -4 sec -1 1.4 x10 -4 / 0.17 = 8.2 x10 -4 sec -1 7.5 x10 -5 M/sec /0.085 M = 8.8 x10 -4 sec -1 Rate is proportional to [H 2 O 2 ] o :

18. Data on how the rate of H 2 O 2 decomposition is affected by varying the Initial [ I -] values. 2X 4.1 X 2X

19. So Rate depends on [H 2 O 2 ] o : Rate rxn = k [H 2 O 2 ] o AND Rate depends on [KI] o : Rate rxn = k* [KI] o Overall, Rate depends on two parameters: Rate rxn = k’ [H 2 O 2 ] o [KI] o where k’= k k* And we say the overall reaction is Second Order, 2 o, First order, 1 o, in H 2 O 2 and First order, 1 o, in KI

20. This expression where both dependences are written: Rate rxn = k’ [H 2 O 2 ] o [KI] o is the Rate law. The Rate Law is the reason Kinetics studies are done: It shows us the slowest step in reaction sequence: the Rate Determining Step, r.d.s.

21. Obtaining Rate Constants from Kinetic Data

22. Examples of Plots of Different Reaction Orders

23. Integrated Rate Laws