1. Chemical Kinetics

2. Kinetics The study of reaction rates. Spontaneous reactions are reactions that will happen - but we can’t tell how fast. (Spontaneity implies nothing about speed) For example: Diamond will spontaneously turn to graphite – so slow it is not detectable. Reaction mechanism- the steps by which a reaction takes place.

3. Reaction Rate Defined as the change in the concentration of a reactant or product per unit time. Rate=Conc.of A at t 2 -Conc. of A at t 1 t 2 – t 1 t 2 – t 1 Rate = [A] D t

4. For this reaction: N 2 + 3 H 2  2 NH 3 As the reaction progresses the concentration H 2 goes down CONCENTRATIONCONCENTRATION TIME [H2]

5. l As the reaction progresses the concentration N 2 goes down 1/3 as fast [H 2 ] [H2] [N2] CONCENTRATIONCONCENTRATION TIME

6. l As the reaction progresses the concentration NH 3 goes up. [H 2 ] [N 2 ] [NH 3 ]

7. Calculating Rates Average rates are taken over long intervals Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time

8. Factors that Affect Reaction Rates Nature of the reactants Concentration of the reactants Temperature Presence of a catalyst

9. Defining Rate We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product. In our example: N 2 + 3H 2  2NH 3 -[N 2 ] = -3[H 2 ] = 2[NH 3 ] t t t

10. Rate Laws Reactions are reversible. As products accumulate they can begin to turn back into reactants. Initially, the rate will depend on only the amount of reactants present. Therefore, reactions must be studied at a point soon after the reactants are mixed. This is called the Initial rate method.

11. Rate Laws The concentration of the products do not appear in the rate law because this is an initial rate. The order must be determined experimentally, and can’t be obtained from the equation

12. 2 NO 2  2 NO + O 2 The rate will only depend on the concentration of the reactants. Rate = k[NO 2 ] n This is called a rate law expression. k is called the rate constant. n is the “order” of the reactant -usually a positive integer and must be determined experimentally Overall reaction order is the sum of the orders for each individual reactant.

13. Types of Rate Laws Differential Rate law - describes how rate depends on concentration (often simply called rate law) Integrated Rate Law - describes how concentration depends on time. For each type of differential rate law there is an integrated rate law and vice versa. Rate laws can help us better understand reaction mechanisms.

14. Determining the form of a rate law (Method of Initial Rates) The initial rate is the instantaneous rate right after the reaction begins (t=0) Eliminates the effect of the reverse reaction. The initial concentrations of the reactants are varied. The reaction rate is determined for each trial. See the example problem on p. 658

15. A + 2B  C Trial Initial [A] Initial [B] Initial Rate of formation of [C] 1 0.010 M 1.5 x 10 -6 M/s 2 0.010 M 0.020 M 3.0 x 10 -6 M/s 3 0.020 M 0.010 M 6.0 x 10 -6 M/s

16. A + B + C  products The rate law is determined to be rate = k[A][B] 2 What happens to the reaction rate when we make the following changes to the concentrations?

17. The concentration of A is doubled while B and C remain unchanged. The concentration of B is doubled while A and C remain unchanged. The concentration of C is doubled while A and B remain unchanged. The concentration of all three reactants are doubled simultaneously.

18. Homework problems to be completed by tomorrow Chapter 16: problems 13-20, 21, 23, 25, 27

19. Collision Theory of Reaction Rates For a reaction to occur, particles must collide. Increasing the concentration or reactants results in a greater number of collisions and therefore, a faster rate. Not all collisions are effective collisions.

20. Effective Collisions For a collision to be effective: 1) particles must possess a necessary minimum energy to break bonds and form new ones (activation energy) 2) particles must have proper orientation at the time of the collision. (see page 677)

21. Reaction Mechanism The step by step pathway by which a reaction occurs is called its mechanism. The individual steps are called elementary steps. The rate law for an elementary step can be written from its molecularity. Molecularity is defined as the number of species that must collide to produce the reaction indicated by that step.

22. Examples of Elementary Steps Elementary Step Molecularity Rate Law A  products Unimolecular Rate = k[A] A + A  products Bimolecular Rate =k[A] 2 A + B  products Bimolecular Rate =k[A][B] A + A + B  products Termolecular Rate=k[A] 2 [B] A + B + C  products TermolecularRate=k[A][B][C]

23. Reaction Mechanism (continued) In most mechanisms, one step is slower than the others. A reaction can never occur faster than its slowest step. The slow step is the rate-determining step.

24. Determining Possible Reaction Mechanisms The balanced equation for the overall reaction is equal to the sum of all of the individual steps. The rate law expression matches the coefficients of the rate determining step.

25. Example NO 2 + CO  NO + CO 2 rate = k[NO 2 ] 2 One proposed mechanism: NO 2 + NO 2  N 2 O 4 (slow) NO 2 + NO 2  N 2 O 4 (slow) N 2 O 4 + CO  NO + CO 2 + NO 2 (fast) N 2 O 4 + CO  NO + CO 2 + NO 2 (fast) NO 2 + CO  NO + CO 2 NO 2 + CO  NO + CO 2 N 2 O 4 is an intermediate (forms in one step and is consumed in a later step) This proposed mechanism is consistent with the experimentally determined rate law.

26. Another possible mechanism NO 2 + NO 2  NO 3 + NO (slow) NO 3 + CO  NO 2 + CO 2 (fast) NO 3 + CO  NO 2 + CO 2 (fast) NO 2 + CO  NO + CO 2 NO 2 + CO  NO + CO 2 In order to be a possible mechanism the following two criteria must be met: 1) The individual steps must add up to the overall reaction 2) The mechanism is consistent with the experimentally determined rate law expression

27. Equilibrium Step An equilibrium step is a step in which a product can rapidly re-form the reactants to reach equilibrium. The concentration of products is equal to the concentration of the reactants at equilibrium. An equilibrium step is indicated by a double arrow. ( ) Substitutions can be made in the rate law expression between the reactants and the products.

28. Example 2NO + Br 2  2NOBr 2NO + Br 2  2NOBr rate = k[NO] 2 [Br 2 ] rate = k[NO] 2 [Br 2 ] Proposed mechanism: NO + Br 2 NOBr 2 (fast) NO + Br 2 NOBr 2 (fast) NOBr 2 + NO  2NOBr (slow) NOBr 2 + NO  2NOBr (slow) 2NO + Br 2  2NOBr 2NO + Br 2  2NOBr Rate = k[NOBr 2 ][NO] can be substituted with Rate =k[NO][Br 2 ][NO] or Rate=k[NO] 2 [Br 2 ]

29. Homework problems to be completed by tomorrow Chapter 16: problems 65, 67, 69, 71, 73, and 75

30. Integrated Rate Laws Integrated Rate Laws express the reactant concentration as a function of time (instead of rate as a function of the reactant concentration). We will only be considering reactions involving a single reactant.

31. First-Order Rate Laws Consider the following reaction: 2N 2 O 5  4NO 2 + O 2 The rate law (differential) is determined to be Rate = k[N 2 O 5 ] This means that if the concentration of N 2 O 5 is doubled, the rate of production of the products is also doubled.

32. Integrated First-Order Rate Law The previous differential rate law can be integrated and as a result, written in the following form: ln[N 2 O 5 ] = -kt + ln[N 2 O 5 ] 0 Where: ln indicates the natural log t is the time [N 2 O 5 ] is the concentration at time “t” [N 2 O 5 ] o is the initial concentration

33. Integrated First-Order Rate Laws All integrated first order rate laws take the following general form: ln[A] = -kt + [A] o Items to note: 1) The equation shows how the concentration of A depends on time 1) The equation shows how the concentration of A depends on time

34. Integrated First-Order Rate Laws ln[A] = -kt + ln[A] o 2) The above equation is of the form y=mx + b, where a plot of y vs. x is a straight line with slope m and intercept b. In the above rate law: In the above rate law: y=ln[A] x=t m=-k b=ln[A] o As a result, plotting ln[A] vs. time always gives a straight line. (This fact is often used to test whether a reaction is 1 st order or not) As a result, plotting ln[A] vs. time always gives a straight line. (This fact is often used to test whether a reaction is 1 st order or not) 3) The integrated rate law for a 1 st order reaction can also be written as: ln([A] o /[A]) = kt

35. Half-Life The time required for a reactant to reach half of its original concentration is called the half-life of a reactant and is designated with the symbol t 1/2. The general equation for the half-life of a first order reaction is t 1/2 =0.693/k (Half-life does not depend on concentration)

36. Second-Order Rate Laws The integrated second order rate law has the form: 1/[A] = kt + 1/[A] o Note: 1) A plot of 1/[A] vs. time produces a straight line with a slope=k. 2) The half-life equation for a 2 nd order reaction is: t 1/2 = 1/k[A] o

37. Zero-Order Rate Laws The rate law for a zero order reaction is: rate = k The integrated rate law for a zero order reaction is [A] = -kt + [A] o Note: 1) A plot of [A] vs. time results is a straight line 2) the half life equation is: t 1/2 = [A] 0 /2k

38. Example Problem 16-10 p. 674 Complete homework problems 33,35,39, and 41. Complete the supplemental problems given out in class today.