1. Chemical Kinetics The “Speed” of the Reaction Or Reaction Rates

2. Reaction Kinetics ReactantsProducts Rate of Change AverageInstantaneous Rate Law Reaction Order Integrated Rate Forms of Rate Law Graphical Analysis of Rates Initial rates Method Half-Life

3. Reactants  What happens to the reactants in a reaction? If we measure the concentration of reactants as a reaction proceeds, what would the graph look like? Or How does the concentration vary with time? –Is it linear? –Exponential? –Random? Do all reactions only move forward? –Assume for now there is no reverse reaction –Or the reverse reaction proceeds so slowly, we are willing to ignore it Map graph

4.  Products Where do the products come from? If we measure the concentration of the products from the first second the reaction starts, how would the concentration vary with time? Map graph

5. Rate of Change In the reaction A  B How does the [A] vary with time? Or What is the rate of the reaction? Rate = [A] time2 – [A] time1 Time 2 – Time 1 [A] = the molarity of A Map

6. Rate of Change The symbol delta, , means “the change in” So the reaction rate can be written Rate = [A] t2 – [A] t1 or time 2 – time 1 Rate =  [A]  t graph

7. Average Rate of Change The concentration varies as time goes on. –Because we use up reactants We can calculate an average rate of product consumption – over a period of time. Rate is like velocity of a reaction. –the rate of change of meters vs rate of change of [A] –Instead of meters per second, it is concentration per second. –To calculate speed/velocity, divide the distance traveled by the time it took. Meters = m 2 – m 1 =  m t 2 – t 1  t sec graph

8. Average Rate of Change What is the change in concentration? (how far did it go? If you are calculating speed) [A] @ time2 – [A] @time1=  [A] How much time has elapsed? [A] @ time2 – [A] @time1=  [A] t 2 – t 1  t graph

9. Instantaneous Rate An average rate describes what reaction rate over a time, but does not tell us the rate at any particular moment. The rate at any moment is the instantaneous rate

10. Instantaneous Rate If we take the average rate over a period of time and continuously make the time period smaller When the time period is infinitesimally small, you approach the instantaneous rate Graphically, it is the slope of the tangent line at the instant.Graphically –That’s why graphing programs have that tangent line function! Rates are important in bio, physics and chem. Map

11. (differential) Rate Law Expresses how rate depends on concentration. Rate = -  [Reactant] = k [reactant] n  t k is the rate constant –The bigger the k value, the faster the reaction –The smaller the k value, ….? n = the order of the reaction and must be determined experimentally. Map

12. Reaction Rates Reaction rates are considered positive Rate instantaneous = k instantaneous = - slope of tangent line Rate average = k average = - ([A] 2 - [A] 1 ) t 2 -t 1 So the rate constant, k, is always negative Rate = -  [Product] = k [Product] n  t –Assuming no reverse reaction!

13. Average Rate of Change From 0 to 300 s = 0.01 –0.0038 = 0.000021 M/s 300 s Product Formation Reaction Rate Average Rate Instantaneous Rate Back to: Reactant/product

14. 2NO 2  2NO + O 2 Let’s consider the above reaction How can we measure the rate? –What data do we need? Measure the time Measure the concentration –We will take advantage of color in our lab –If we are measuring light, we are doing….. SPECTROSCOPY

15. Spectrometer Source Monochrometer, LED Or Filter Sample Detector The sample absorbs the light. The detector determines how much. Many frequencies of light One frequency Of light

16. Beer’s Law Beer’s law states that the amount of light absorbed depends on: –The material molar absorbtivity (physical property) –How much is there? molarity –And how big the sample holder The light spends more “time” in contact with a longer sample

17. Spectrometer Source Monochrometer, LED Or Filter Sample Detector The sample absorbs the light. The detector determines how much. Many frequencies of light One frequency Of light

18. Spectroscopy Assume the concentration is directly proportional to absorbance of light The more stuff there is that absorbs the light – the less light that goes through …. or –More light is absorbed Beer’s Law a = e l c = k M a = absorbtivity e = molar absorbtivity (physical property) l = length of light path c = molarity or the solution Molarity and absorbtivity Are directly proportional

19. 2NO 2  2NO + O 2 Time[NO 2 ][NO][O 2 ] 00.010000 500.00790.0210.0011 1000.00650.00350.0018 1500.00550.00450.0023 2000.00480.00520.0026 2500.00430.00570.0029 3000.00380.00620.0031 3500.00340.00660.0033 4000.00310.00690.0035 Compare the  [NO 2 ] in the first 50 secs and the last 50 secs Why does the rate slow down?

20. Formation of Products 2NO 2  2NO + O 2 Rate of Consumption NO 2 = Rate Formation NO Rate = k[NO 2 ] = - k[NO] Because For every two NO 2 consumed two NO formed

21. Formation of Products 2NO 2  2NO + O 2 Rate of Consumption NO 2 = 2 x Rate Formation O 2 Rate = k[NO 2 ] = - k/2 [O 2 ] Because For every two NO 2 consumed one O 2 formed

22. Compare the Instantaneous Rates At any moment in time  [NO 2 ] = -  [NO] = 2 -  [O 2 ]  t  t  t Or k [NO 2 ] = - k [NO] = - k/2 [O 2 ] graph

23. Form of the Rate Law For aA + bB  cC +dD Rate = k [A] n [B] m –Where k is the rate constant n = order of reactant A m = order of reactant B n and m must be determined experimentally n +m = order of the reaction

24. Experimental Order the order in the integrated rate law Rate = -  [Reactant] = k [Reactant] n  t n = 0, zero order n = 1, first order n = 2, second order Determine order

25. Order of Reaction A + B → C Rate = k[A] n [B] m (n + m) = order of the reaction = 1 unimolecular =2 bimolecular =3 trimolecular This means how many particles are involved in the rate determining step

26. Method of Initial Rates A series of experiments are run to determine the order of a reactant. The reaction rate at the beginning of the reaction and the concentration are measured These are evaluated to determine the order of each reactant and the overall reaction order

27. If you plot the concentration versus time of [N 2 O 5 ], you can determine the rate at 0.90M and 0.45M. What is the rate law for this reaction? Rate = k [N 2 O 5 ] n n = the order. It is determined experimentally.

28. 2N 2 O 5(soln)  4NO 2(soln) + O 2(g) At 45  C, O 2 bubbles out of solution, so only the forward reaction occurs. Data [N 2 O 5 ]Rate ( mol/l s) 0.90M5.4 x 10 -4 0 45M2.7 x 10 -4 The concentration is halved, so the rate is halved

29. 2N 2 O 5(soln)  4NO 2(soln) + O 2(g) Rate = k [N 2 O 5 ] n 5.4 x 10 -4 = k [0.90] n 2.7 x 10 -4 = k [0 45] n after algebra 2 =(2) n n = 1 which is determined by the experiment Rate = k [N 2 O 5 ] 1

30. Method of Initial Rates Measure the rate of reaction as close to t = 0 as you can get. This is the initial rate. Vary the concentration Compare the initial rates. Map

31. NH 4 + + NO 2 -  N 2 + 2H 2 O Rate = k[NH 4 +1 ] n [NO 2 -1 ] m How can we determine n and m? (order) Run a series of reactions under identical conditions. Varying only the concentration of one reactant. Compare the results and determine the order of each reactant Order

32. NH 4 + + NO 2 -  N 2 + 2H 2 O Experiment[NH 4 ] + Initial [NO 2 ] - Initial Initial Rate Mol/L ·s 10.001M0.0050 M1.35 x 10 -7 20.001M0.010 M2.70 x 10 -7 30.002M0.010M5.40 x 10 -7

33. NH 4 + + NO 2 -  N 2 + 2H 2 O Compare one reaction to the next 1.35 x 10 -7 = k(.001) n (0.050) m 2.70 x 10 -7 = k (0.001) n (0.010) m Exp[NH 4 ] + Initial [NO 2 ] - Initial Initial Rate Mol/L ·s 10.001M0.0050 M1.35 x 10 -7 20.001M0.010 M2.70 x 10 -7 30.002M0.010M5.40 x 10 -7 Form

34. 1.35 x 10 -7 = k(0.001) n (0.0050) m 2.70 x 10 -7 k (0.001) n (0.010) m In order to find n, we can do the same type of math with the second set of reactions 1.35 x 10 -7 = (0.0050) m 2.70 x 10 -7 (0.010) m 1/2 = (1/2) m m = 1

35. NH 4 + + NO 2 -  N 2 + 2H 2 O Compare one reaction to the next 2.70 x 10 -7 = k (0.001) n (0.010) m 5.40 x 10 -7 = k(.002) n (0.010) m Exp[NH 4 ] + Initial [NO 2 ] - Initial Initial Rate Mol/L ·s 10.001M0.0050 M1.35 x 10 -7 20.001M0.010 M2.70 x 10 -7 30.002M0.010M5.40 x 10 -7

36. 2.70 x 10 -7 = k (0.001) n (0.010) m 5.40 x 10 -7 k(.002) n (0.010) m n + m = order of the reaction 1 + 1 = 2 or second order Form 0.5 = (0.5) n n = 1

37. Review Method of initial rate In the form Rate = k[A] n [B] m –Where k is the rate constant – n, m = the order of the reactant The order is determined experimentally Rate law is important so we can gain an insight into the individual steps of the reaction

38. The Integrated Rate Law Expresses how concentrations depend on time Depends on the order of the reaction Remember Rate = k[A] n [B] m Order = n + m Integrated Rate law takes the form by “integrating” the rate function. (calculus used to determine) –The value of n and m change the order of the reaction –The form of the integrated rate depends on the value of n –You get a different equation for zero, first and second order equations. Map

39. Reaction Order Order of the reaction determines or affects our calculations. Zero order indicates the use of a catalyst or enzyme. The surface area of catalyst is the rate determining factor. First or second order is more typical (of college problems)

40. Integrated Law - Zero Order Rate = -  [A] = k  t Set up the differential equation d[A] = -kt Integral of 1 with respect to A is [A]

41. Integrated Rate Law – First Order Rate =  [A] = k [A] n  t If n = 1, this is a first order reaction. If we “integrate” this equation we get a new form. Ln[A] = -kt + ln[A 0 ] where A 0 is the initial concentration Map

42. Why? If Rate = -  [A] = k [A] 1  t Then you set up the differential equation: d[A] = -kdt [A] Integral of 1/[A] with respect to [A] is the ln[A].

43. Integrated Rate Law ln[A] = -kt + ln[A] 0 The equation shows the [A] depends on time If you know k and A 0, you can calculate the concentration at any time. Is in the form y = mx +b Y = ln[A] m = -k b = ln[A] 0 Can be rewritten ln( [A] 0 /[A] ) = kt This equation is only good for first order reactions!

44. First Order Reaction [N 2 O 5 ]Time (s) 0.1000 0 0.0707 50 0.0500100 0.0250200 0.0125300 0.00625400 Ln[N 2 O 5 ] Time (s)

45. ZeroFirstSecond Rate Law Rate = K[A] 0 Rate = K[A] 1 Rate = K[A] 2 Integrated Rate Law [A] = -kt + [A] 0 Ln[A] = -kt +ln[A] 0 1 = kt + 1 [A] [A] 0 Line [A] vs t ln[A] vs t 1 vs t [A] Slope = - k k Half-life t 1/2 = [A] 0 2k t 1/2 = 0.693 k T 1/2 = 1 k[A] 0 Graph DataMap

46. Given the Reaction 2C 4 H 6  C 8 H 12 [C 4 H 6 ] mol/LTime (± 1 s) 0.010000 0.006251000 0.004761800 0.003702800 0.003133600 0.002704400 0.002415200 0.002086200 And the data

47. 2C 4 H 6  C 8 H 12 Equations

48. Graphical Analysis Ln [C 4 H 6 ] ___1___ [C 4 H 6 ] Data Map

49. Experimental Derivation of Reaction Order Arrange data in the form 1/[A] or ln [A] or [A] Plot the data vs time Choose the straight line y = mx + b Determine the k value from the slope Graphical rate laws 1/[A] = kt + b → 2nd ln[A] = kt + b → 1st [A] = kt + b → zero

50. Half-life The time it takes 1/2 of the reactant to be consumed This can be determined –Graphically –Calculate from the integrated rate law

51. Half-Life Graphical Determination

52. Half-Life Algebraic Determination Half-life t 1/2 = [A] 0 2k t 1/2 = 0.693 k T 1/2 = 1 k[A] 0 Equations are derived from the Integrated Rate Laws. ZeroFirstSecond Map

53. Other Subjects in Kinetics Mechanisms Half lives Activation energy Catalyst