Chapter 12 Chemical Kinetics

1. Chapter 12Chemical Kinetics The study of reactions rates Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed?

2. Chemical Kinetics The study of reaction rates Spontaneous reactions---is the tendency for the reaction to occur but it does not imply how fast. Ex. Diamond will spontaneously turn into graphite --- eventually

3. Reaction Mechanism A goal of chemical kinetics is to understand the steps by which a reaction takes place. The series of steps is called the reaction mechanism. If we understand the mechanism, we can find ways to facilitate the reaction

4. Let’s review collision theory Particles have to collide to react They have to hit hard enough Things that increase this increase rate High temps – faster reaction High concentration – faster reaction Small particles = greater surface area – faster reaction

5. Reaction Rate The brackets mean concentration (Molarity) • Change in concentration of a reactant or product per unit time

6. A B rate = D[A] D[B] rate = - Dt Dt

7. Think about the reaction N2 + 3H2 2NH3

8. Concentration N2 + 3H2→ 2NH3 • As the reaction progresses the concentration H2 goes down [H2] Time

9. Concentration N2 + 3H2→ 2NH3 • As the reaction progresses the concentration N2 goes down 1/3 as fast [N2] [H2] Time

10. Concentration N2 + 3H2→ 2NH3 • As the reaction progresses the concentration NH3 goes up 2/3 times [N2] [H2] [NH3] Time

11. 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 • Derivative.

12. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) slope of tangent slope of tangent slope of tangent [Br2]final – [Br2]initial D[Br2] average rate = - = - Dt tfinal - tinitial instantaneous rate = rate for specific instance in time

13. 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 • N2 + 3H2 2NH3 • D[H2] = 3D[N2] DtDt • D[NH3] = -2D[N2] DtDt

14. 2NO2(g) 2NO(g) + O2 (g) Let’s look at this in terms of numbers & instantaneous rate

15. Rate of NO2 • Since the concentration of the reactant will typically be decreasing over time, the rate expression will include negative; however, the slope of the line will be negative which will then yield a positive rate.

16. Rate of NO 2NO2 (g)  2NO(g) + O2 (g)

17. Rate of O2 2NO2 (g) 2NO(g) + O2 (g)

18. Compare the three rates 2NO2 (g)  2NO(g) + O2 (g) Rate of NO2 = 2.4 x 10-5 mol/L•s Rate of NO = 8.6 x 10-6 mol/L•s Rate of O2 = 4.3 x 10-6 mol/L•s Notice the rate of NO is twice that of O2. This makes sense since twice as much NO is produced according to the reaction.

19. 12.2 Rate Laws: An Introduction We looked at the reaction 2NO2 (g) 2NO(g) + O2 (g) But we need to understand that the reverse happens too. 2NO(g) + O2 (g)  2NO2 (g) Chemical reactions ARE reversible

20. Chemical equilibrium When forward and reverse reaction rates are equal, it is called chemical equilibrium. When this occurs, there will be NO changes in the concentrations of reactants and products.

21. 2NO2 (g) 2NO(g) + O2 (g) in the reaction depends on the difference in the rates of the forward and reverse reactions. If you choose conditions where the reverse reaction can be neglected. Then the reaction rate will depend only on the concentrations of the reactants. The rate for the decomprxn is

22. This shows that the rate depends on the concentration of the reactants…this is the rate law expression. k = the rate constant n = order of the reactant both must be determined by experiment (although n can be an integer or a fraction, we will typically look at positive integer orders in this book.)

23. Two important things: • The concentration of the products is not considered here because the reverse reaction does not contribute to the rate • The value of the exponent n must be determined from experimentation

24. So what do we choose? The question is which reactant or product do we choose for defining the “rate” in the rate equation: The value of the rate constant, k, depends on how the rate is defined.

25. Two types of rate laws One type: Differential rate laws – once you know the value of n – the rate law expresses how the rate depends on concentration. Often just called “the rate law” – rate as a function of concentration.

26. The other type: Integrated rate law – expresses how concentration depends on time Once you find one experimentally, you know the other. The one you find experimentally is based on which is easier to collect data.

27. Why? • What is the purpose of knowing the rate? • Infer the steps by which a reaction occurs • Understand the reaction by learning what the steps are in the reaction • Where can I interrupt the reaction? • How can I speed up the reaction? • Rate laws are not used for the purpose of knowing the rate, but what the rate reveals about the STEPS of the reaction. (Reaction mechanism)

28. 12.3 Determining the Form of the Rate Law aA + bBcC + dD Rate = k [A]x[B]y Reaction is xth order in A Reaction is yth order in B Reaction is (x +y)th order overall The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some power.

29. Lets look at the decomp of N2O5 first. (differential rate) 2N2O5 4NO2 + O2 Rate = k [N2O5]n Here’s some data

30. Notice that the concentration of N2O5 is halved, the rate is also halved. This means that the rate is completely dependent on concentration of N2O5 to the FIRST power. Notice: It is NOT based on the coefficient from the reaction.

31. Initial rates. The initial rate is the instantaneous rate found immediately after the reaction has begun. The experiment must be run several times with various initial concentrations The initial rate is found for EACH “run” Then compare rates to determine the form of the rate law.

32. The reaction:NH4+ + NO2- N2 + 2H2O NO2- The general form of the rate law We need experimental data to determine the values of n and m

33. Simplify NO2- Compare: [NO2-] doubles as the rate doubles there fore m = 1 Here’s the math as a ratio to find m This means that “m” must be 1 Do this again for “n” with rates 2 & 3

34. Simplify NO2- Again, when the [NH4+] doubles so does the rate n = 1 Which mean “n” must be 1 and the rate law is

35. The rate law is first order for both reactants with an overall reaction order of 2 The reaction order is the sum of the orders for the various reactants With all this info you can now calculate the rate constant for this reaction.

36. NO2- Pick ANY one of the experiments and substitute and solve for k. We’ll do the same one as the book. But you could choose either!

37. Use the reaction and data below to find: The orders of each reactant, the overall reaction order, and the value of the rate constant.

38. 12.4 Integrated rate law So far we’ve looked at rate as a function of reactant concentrations. Integrated rate laws relate concentration to reaction time. Let’s look at reactions with a single reactant.

39. aA products What is the basic rate law? First Order - double the concentration… double the rate rate law integrated if n=1

40. Shows how concentration of A depends on time. If [A]o &k are known [A] at any time can be calculated. If plotting ln[A] vs. time gives a straight line where k is the slope, then it proves the reaction is first order. Often shown as a ratio.

41. Half Life Half life is the time required for a reactant to reach half its original concentration. For a first order reaction

42. Second order Starting with the basic rate law Double the concentration…quadruple the rat OR triple the concentration …nine times the rate. Integrated law when n=2

43. If you plot versus t and get a straight line with a slope of k…the reaction is proven to be second order.

44. Half life for second order The equation for 2nd order half life For second order reactions, t1/2 is dependent on [A]0. For first order reactions t1/2 is independent on [A]0. For a 2nd order reaction each successive ½ life is double the preceding one.(ex. The 1st ½ life of C4H6 reacting to form a dimer is 1630 sec, but the 2nd ½ life takes 3570 seconds.)

45. Zero order rate laws The reaction has a constant rate. When n=0 Rate=k Integrated rate law when n=0

46. Zero order half life Most zero order reactions are when a substance such as a metal or an enzyme is required for the reaction to occur.

47. Integrated rate laws for reactions with more than one reactant Most often the kinetics of a complicated reaction can be studied by opbserving the behavior of one reactant at a time. If one reactant has a concentration that is much smaller than the others, then it is used to determine the order of the reaction in that component.

48. Summary

49. Summary • We only consider only the forward reaction so that rate laws contain only reactant concentrations. • There are two types of rate laws. • Differential – rate depends on concentrations. • Integrated – rate is a function of concentration with respect to time.

50. Summary The type of rate law depends on the type of data that is most conveniently collected. Once we have one type of rate law we can determine the other. The most common method for differential rate law is the method of initial rates. For integrated rate laws concentrations are measured at various values of t.