2. Chemical Kinetics A Study of the Rates of Reactions

3. Consider a reaction in which D  A + B How “fast” does the reactant, D, disappear?

4. Concentration of D over Time This is the raw data collected during the experiment Start your analysis by graphing [D] v. Time

7. The data do not form a straight line. The R 2 value only has one “9”. This reaction is NOT zero order! How do we know this? The plot is NOT a straight line! R 2 = 0.9737

8. Next, convert your raw data to natural logs and graph ln[D] v. time

9. What is the difference between a natural log and a common log? Natural logs use a base value of “e”, 2.71 and is designated with the symbol “ln” Common logs use a base value of “10” and is designated with the symbol “log” The conversion between natural and common logs is: log x = 2.303 ln x

12. This reaction is first order. Ln [D] time The R2 value is 1.0. The data DO form a straight line. Plot of ln [D] vs time

13. Normally, when you get a straight line you stop making graphs. For teaching purposes, let’s go ahead and look at the third kind of graph: 1/[D] v. time

16. Reaction Rates average rate the rate over a specific time interval instantaneous rate the rate for an infinitely small interval

17. Rate Law Reaction rate = k [A] m [B] n wherem => order with respect to A n => order with respect to B overall order = m + n

18. Order of Reaction exponent of the concentration for a reactant that implies the number of molecules of that species involved in the rate determining step first order, exponent equals one second order, exponent equals two

19. Rates are studied for the first few moments of the chemical reaction.

20. Concentrations are changed and the change in rate is calculated.

21. The initial rate is found by calculating the AVERAGE RATE over the first few moments of the reaction.

22. We know that the average rates will be different because the slope of each line is different.

23. When several initial concentrations have been tried, we can generate a table of data that looks like this: 2R  P

24. What would the basic rate law look like for this reaction? 2R  P Rate = k [R] m The equation says that the concentration of R is proportional to the rate of reaction. [R] m : rate

25. We need to solve for m and k. We start by solving for m. Basic Rate Law:Rate = k [R] m Chemical Reaction: 2R  P

26. Solving for m using the Method of Initial Rates Rate = k [R] m

27. Basic Rate Law:Rate = k [R] m Chemical Reaction: 2R  P By what factor did the concentration increase from experiment 1 to experiment 2? It doubled. Experiment 2 = 0.2 = 2 Experiment 1 0.1

28. Basic Rate Law:Rate = k [R] m This is a 1:1 relationship between concentration and rate. How can we use the Basic Rate Law to guide us to this relationship? [R] m : rate The rate goes up when the concentration goes up. 2 m = 2 The concentration doubled and the rate doubled. m = 1 Solve for m Thus, the rate law is: Rate = k [R] 1 or Rate = k [R]

29. Let’s review some basic math: 2 0 = 2 1 = 2 2 = 2 3 = 2 4 = 4 2 1 8 16 3 m = 9 3 m = 3 4 m = 16 5 m = 125 456 m = 1 m = 2 m = 1 m = 2 m = 3 m = 0

30. Solving for k using the Method of Initial Rates Rate = k [R] m

31. The Rate Law is :Rate = k [R] Now, we can solve for k by inserting the information in the table into the Rate Law. Rate = k [R] 2.7 x 10 -5 M/s = k [0.1 M] k = 2.7 x 10 -4 s -1

32. Experiment 1: k = 2.7 x 10 -4 s -1 Experiment 2: k = 2.7 x 10 -4 s -1 Experiment 3: k = 2.7 x 10 -4 s -1 The value for k, the proportionality constant, is the same for all three experiment. It is a good idea to check at least two experiments to see that k is the same. If you made a mistake finding the rate law, the k ’s will differ!

33. We have found 1. The rate law for the reaction R  P: Rate = k [R] 2. The value of the proportionality constant, k : k = 2.7 x 10 -4

34. Now we can write the rate law for the reaction R  P Rate = (2.7 x 10 -4 )[R] Final form for the Rate Law for this example:

35. Average Rates Chemical Kinetics of NO 2

39. Rate =  [NO 2 ] = 0.0031 mol/L - 0.0100 mol/L = -1.73 x 10 -5 mol/L s  time 400 s - 0 s Why is the rate negative? How do we fix this problem? Rate = -  [NO 2 ] = - (0.0031 mol/L - 0.0100 mol/L) = 1.73 x 10 -5 mol/L s  time 400 s - 0 s We add a negative sign when a chemical is disappearing!

40. For all reactants: Rate = -  [reactant]  time For all products: Rate =  [product] D time Use a negative sign for reactant rates! Omit the negative sign for product rates!

42. Advantages of the Method of Initial Rate: 1. Useful when a reaction is reversible. The reverse reaction won’t significantly contribute to the first few moments of the reaction. 2. Useful for very fast or very slow reactions.

43. Advantages of the Integrated Rate Method: 1. Useful for moderate length reaction. 2. Doesn’t require multiple experiments to determine the order of the reaction.

44. Common Uses of the Method of Initial Rates: 1. To determine the order of the reaction. 2. Find the rate constant, k.

45. Common Uses of the Method of Integrated Rate Laws: 1. To determine the order of the reaction. 2. Find the rate constant, k. 3. To determine the concentration at a certain time. 4. To determine at what time a certain concentration will be reached.

46. Integrated Rate Laws A ---> products rate = - (  [A]/  t) = k[A] m average rate rate = - (d[A]/dt) = k[A] m instantaneous rate

48. Integrated Rate Law: A reaction is followed for an extended period of time. Method of Initial Rates A reaction is followed for only the first few moments of the reaction.