AP Notes Chapter 15

1. AP Notes Chapter 15 Principles of Reactivity: Chemical Kinetics

2. Chemical Kinetics • The study of how fast chemical reactions occur. • How fast species are appearing and disappearing. • It has nothing to do with the extent to which a reaction will occur – that is thermodynamics.

3. There is a difference between the probability that molecules will react when they meet each other and once they do react how stable is the product. • First, probability of reaction has to do with overcoming an energy barrier or activation energy. • Second, has to do with the differences between reactants and products.

4. Thermodynamics: • Spontaneity deals with how likely a reaction will take place. • Spontaneous reactions are reactions that will happen - but we can’t tell how fast. • Diamond will spontaneously turn to graphite – eventually.

5. Factors Affecting Rates 1. Surface area Powders 2. Temperature Dye 3. Pressure 4. Catalyst /Inhibitor 5. Concentration H202 & I2

6. Reaction Rate The change in concentration of a reactant or product per unit time For a reaction A  B Average Rate = Change in Moles of B Change in time Rate = D [B] = - D [A] D t D t

7. Reaction Rate • Rate = Conc. of A at t2 -Conc. of A at t1t2- t1 • Rate =D[A]Dt • Change in concentration per unit time • Changes in Pressure  to changes in concentration

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. [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 and Integrated Rates.

12. Concentration • Average slope method D[H2] Dt Time

13. Concentration • Instantaneous slope method. D[H2] D t Time

15. 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[N2] = - D[H2] = D[NH3] Dt 3 Dt 2 Dt

16. RATE LAW Mathematical model that contains specific information about what reactants determine the rate and how the rate is controlled

17. Rate Laws • Reactions are reversible. • As products accumulate they can begin to turn back into reactants. • Early on the rate will depend on only the amount of reactants present. • We want to measure the reactants as soon as they are mixed. • This is called the Initial rate method.

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

19. 2 NO2 2NO + O2 • You will find that the rate will only depend on the concentration of the reactants. • Rate = k[NO2]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.

20. 2 NO2 2NO + O2 • The rate of appearance of O2 can be said to be. • Rate' = D[O2] = D[NO2] = k'[NO2]Dt 2Dt • Because there are 2 NO2 for each O2 • Rate = 2 x Rate' • So k[NO2]n= 2 x k'[NO2]n • So k = 2 x k'

21. Types of Rate Laws • Differential Rate law - describes how rate depends on concentration. • 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.

22. Determining Rate Laws • The first step is to determine the form of the rate law (especially its order). • Must be determined from experimental data. • For this reaction 2 N2O5 (aq) 4NO2 (aq) + O2(g)The reverse reaction won’t play a role

23. [N2O5] (mol/L) Time (s) 1.00 0 0.88 200 0.78 400 0.69 600 0.61 800 0.54 1000 0.48 1200 0.43 1400 0.38 1600 0.34 1800 0.30 2000 Now graph the data

24. To find rate we have to find the slope at two points • We will use the tangent method.

25. At .90 M the rate is (.98 - .76) = 0.22 =- 5.5x 10 -4 (0-400) -400

26. At .40 M the rate is (.52 - .31) = 0.22 =- 2.7 x 10 -4 (1000-1800) -800

27. Since the rate at twice the concentration is twice as fast the rate law must be.. • Rate = -D[N2O5] = k[N2O5]1 = k[N2O5] Dt • We say this reaction is first order in N2O5 • The only way to determine order is to run the experiment.

28. The method of Initial Rates • This method requires that a reaction be run several times. • The initial concentrations of the reactants are varied. • The reaction rate is measured bust after the reactants are mixed. • Eliminates the effect of the reverse reaction.

29. Integrated Rate Law • Expresses the reaction concentration as a function of time. • Form of the equation depends on the order of the rate law (differential). • Rate = D[A]nDt • We will only work with n=0, 1, and 2 although negative and fractional exist

30. General Rate Law aA + bB  cC Rate  -[A] or -[B] r = k[A]m[B]n • Where: • m = order with respect to A • n = order wrt B • and n + m = order of reaction Notice we are primarily discussing the reactants

31. For example: aA + bB  cC + dD • Zero Order: r = k • First Order in A: r = k [A] • First Order in B: r = k [B] • Second Order in A: r = k [A]2 • Second Order in B: r = k [B]2 • First Order in A, first Order in B, for a second order overall: r = k [A][B]

32. Initial Rate Method to establish the exact rate law R0 = k[A0]m[B0]n but experimental data must provide orders Use ratio method to solve for m and n R0 = k[A0]m[B0]n

33. R0 = k[A0]m[B0]n Exp Rate [A0] [B0] 1 0.05 0.50 0.20 2 0.05 0.75 0.20 3 0.10 0.50 0.40 Thus, rate = k[A]0[B]1 or rate = k[B]

34. Where k is the rate constant , with appropriate units. rate = k[B] Using Exp 1:

35. INTEGRATED RATE LAWS Each “order” has a separate “law” based on the integration of the rate expression.

36. You NEED TO KNOW the rate expressions & half-life formulas for 0, 1st, and 2nd order reactions. Third order and fractional order reactions exist, but are rare. Nuclear reactions are typically 1st order, therefore, use the first-order rate law and half-life to solve problems of this type

38. Zero Order Rate Law • Most often when reaction happens on a surface because the surface area stays constant. • Also applies to enzyme chemistry.

39. Concentration Time

40. Concentration k = D[A]/Dt D[A] Dt Time

41. Zero Order Rate Law • Rate = k[A]0 = k • Rate does not change with concentration. • Integrated [A] = -kt + [A]0 • When [A] = [A]0 /2 t = t1/2 • t1/2 = [A]0 /2k

42. For a zero-order reaction, the rate is independent of the concentration. Expressed in terms of calculus, a zero - order reaction would be expressed as:

43. m = -k [B] t

44. Time for the concentration to become one-half of its initial value Half-life 0-Order Reaction

45. First Order • For the reaction 2N2O5 4NO2 + O2 • We found the Rate = k[N2O5]1 • If concentration doubles rate doubles. • If we integrate this equation with respect to time we get the Integrated Rate Law • ln[N2O5] = - kt + ln[N2O5]0 • ln is the natural log • [N2O5]0 is the initial concentration.

46. First Order • General form Rate = D[A] / Dt = k[A] • ln[A] = - kt + ln[A]0 • In the form y = mx + b • y = ln[A] m = -k • x = t b = ln[A]0 • A graph of ln[A] vs time is a straight line.

47. First Order • By getting the straight line you can prove it is first order • Often expressed in a ratio

48. First Order • By getting the straight line you can prove it is first order • Often expressed in a ratio

49. First OrderNuclear Decay and Half life • The time required to reach half the original concentration. • If the reaction is first order • [A] = [A]0/2 when t = t1/2

50. The time required to reach half the original concentration. • If the reaction is first order • [A] = [A]0/2 when t = t1/2 • ln(2) = kt1/2