Chapter 12: Chemical Kinetics

1. Chapter 12:Chemical Kinetics • Spontaneity – inherent tendency for a reaction to occur; does not mean speed • Reactions have different rates: • 2 H2 + O2 2 H2O SLOW • Explosions FAST • To be useful, reactions must occur at a reasonable rate.

2. Chemical Kinetics • The area of chemistry concerned with the speed (rate) of a reaction and reaction mechanisms.

3. Main Goal of Kinetics • Understand the steps by which a reaction takes place • What factors determine how fast food spoils? • Design of fast setting material for dental fillings • What affects the rate that steel rusts? • What affects the rate the fuel burns in cars?

4. Reaction Rate Factors • Physical state of reactants • Concentrations of reactants • Temperature • Presence of catalyst • Force of collisions

5. Reaction Rate • Change in concentration(conc) of a reactant or product per unit time. (where A is a reactant or product) (note that rate is always a + value) Unit typically M/s

6. Progress of a hypothetical rxn

7. Sample Exercise 1 • Using the data given in slide 6, calculate the average rate at which A disappears over the time interval 20 s to 40 s.

8. 2 NO2(g) 2 NO(g) + O2(g)

9. Find the avg rate the first 50 s of NO2 change. We want to work with positive numbers so the equation will be:

10. Concentration of nitrogen dioxide, nitric oxide, and oxygen versus time.

11. What do we discover about the rate of this reaction by looking at the graph on slide 10? • Instantaneous rate – value of the rate at particular time; found by calculating the slope of a line tangent to the curve at that point • At 100 s:

12. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • All reactions slow down over time. • Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction. (called initial rate)

13. Using the graph on slide 12, calculate the instantaneous rate of disappearance of reactant at t = 0.

14. Rate Laws Rate = k[A]m [B]n k = rate constant, dependent on temperature m & n = reaction orders

15. Rate Laws The rate law for any reaction must be determined experimentally, it cannot be predicted by merely looking at the chemical equation. *usually involves only the concentration of reactants

16. Types of Rate Laws • Differential Rate Law: expresses how rate depends on concentration. (Often just called “rate law”) • Integrated Rate Law: expresses how concentration depends on time. • They are inter-related, so once you experimentally determine one, the other can be found.

17. WHY???!!!! • We can work backward from the rate law to infer the steps by which the reaction occurs which helps us to find the slowest step – then we can figure out how to speed it up!

18. Method of Initial Rates • Initial Rate: the “instantaneous rate” just after the reaction begins (just after t = 0) • The initial rate is determined in several experiments using different initial concentrations.

19. N2(g) + 2 H2O(l) NH4+(aq) + NO2−(aq) Method of Initial Rates: Sample 1 Rate = k [NH4+]m[NO2−]n

20. Method of Initial Rates: Sample 1 Determine the values of n and m. The value of m is 1

21. Method of Initial Rates: Sample 1 A similar method is used to find n. The value of n is also 1.

22. Method of Initial Rates: Sample 1 The values of n and m are both 1 and the rate law is: This rate law is first order in both NO2- and NH4+ . The overall reaction order is the sum of n and m. The reaction is second order overall.

23. For the reaction • NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l) • we can observe that • as [NH4+] doubles with [NO2-] constant the rate doubles, • as [NO2-] doubles with [NH4+] constant, the rate doubles, • We conclude rate  [NH4+][NO2-]. • Rate law: • The constant k is the rate constant.

24. Calculate the rate constant k Then

25. Using Initial Rates to Determine Rate Laws • A reaction is zero order in a reactant if the change in concentration of that reactant produces no effect. • A reaction is first order if doubling the concentration causes the rate to double. • A reaction is nth order if doubling the concentration causes an 2n increase in rate. • Note that the rate constant does not depend on concentration.

26. Overall Reaction Order • Sum of the order of each component in the rate law. • rate = k[H2SeO3][H+]2[I-]3 • The overall reaction order is 1 + 2 + 3 = 6.

27. Kinetic Sample Problem 1 2 H2 + 2 NO 2H2O + N2 at 800 K

28. Complete Sample Problems 2-4

29. = −kt ln [A]t [A]0 Integrated Rate Laws Using calculus to integrate the rate law for a first-order process gives us Where [A]0 is the initial concentration of A, and [A]t is the concentration of A at some time, t, during the course of the reaction.

30. = −kt ln [A]t [A]0 Integrated Rate Laws Manipulating this equation produces… ln [A]t− ln [A]0 = − kt On purple equation sheet ln [A]t = − kt + ln [A]0 …which is in the form y = mx + b

31. First-Order Processes Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k. ln [A]t = -kt + ln [A]0

32. CH3NC CH3CN First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile.

33. CH3NC CH3CN First-Order Processes This data was collected for this reaction at 198.9 °C.

34. First-Order Processes • When ln P is plotted as a function of time, a straight line results. • Therefore, • The process is first-order. • k is the negative of the slope: 5.1  10-5 s−1.

35. If you know the initial concentration and k, the concentration at any time can be calculated. • Example: The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr-1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 x 10-7 g/cm3. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the concentration of the insecticide to decrease to 3.0 x 10-7 g/cm3?

36. Complete sample problems on wkst.

37. 1 [A]0 1 [A]t = kt + Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, we get On purple equation sheet Rearranged to give also in the form y = mx + b

38. 1 [A]0 1 [A]t = kt + 1 [A] Second-Order Processes So if a process is second-order in A, a plot of vs. t will yield a straight line, and the slope of that line is k.

39. NO2(g) NO (g) + O2(g) 1 2 Second-Order Processes The decomposition of NO2 at 300°C is described by the equation and yields data comparable to this:

40. Second-Order Processes • Plotting ln [NO2] vs.t yields the graph below. • The plot is not a straight line, so the process is not first-order in [A].

41. 1 [NO2] Second-Order Processes • Graphing vs. t, however, gives this plot. • Because this is a straight line, the process is second-order in [A].

42. Half-Life • Half-life is defined as the time required for one-half of a reactant to react. • Because [A] at t1/2 is one-half of the original [A], [A]t = 0.5 [A]0.

43. 0.5 [A]0 [A]0 ln = −kt1/2 0.693 k = t1/2 Half-Life For a first-order process, this becomes ln 0.5 = −kt1/2 −0.693 = −kt1/2 NOTE: For a first-order process, then, the half-life does not depend on [A]0.

44. 1 k[A]0 1 [A]0 1 [A]0 1 [A]0 2 [A]0 1 0.5 [A]0 2 − 1 [A]0 = kt1/2 + = kt1/2 + = = kt1/2 = t1/2 Half-Life For a second-order process, NOTE: For a second order reaction, The half-life is dependent upon the initial concentration.

45. Testing for a Rate Law Plot [A] vs t. Plot ln[A] vs t. Plot 1/[A] vs t.

46. Summary of the Kinetics for Reactions that are Zero, First, and Second Order in [A]

47. Temperature and Rate • Most reactions speed up as temperature increases. (E.g. food spoils when not refrigerated.) • Two light sticks are placed in water, one at room temp. and one in ice • the one at room temp. is brighter than the one in ice. • The chemical reaction, responsible for chemiluminescence is dependent on temp. • The higher the temp., the faster the reaction and the brighter the light. • As temperature increases, the rate increases.

48. Collision Model • Key Idea: Molecules must collide to react. • However, only a small fraction of collisions produces a reaction. Why? • Arrhenius: An activation energymust be overcome.

49. The Collision Model • The more molecules present, the greater the probability of collision and the faster the rate. • The higher the temperature, the more energy avail. to the molecules and the faster the rate.

50. The Collision Model • Complication: not all collisions lead to products. In fact, only a small fraction of collisions lead to product. • In order for reaction to occur the reactant molec. must collide in the correct orientation and with enough energy to form products.