1. PRT 140 PHYSICAL CHEMISTRY PROGRAMME INDUSTRIAL CHEMICAL PROCESS SEM 1 2013/2014 CHEMICAL KINETIC BY PN ROZAINI ABDULLAH SCHOOL OF BIOPROSES ENGINEERING ©RBA FTK 2013

2. SUBTOPICS Experimental Chemical and Kinetics Reactions First Order Reactions Second Order Reactions Reaction Rates and Reaction Mechanisms Light Spectroscopy and Adsorption Chemistry (Experimental methods for fast reactions).

3. KINETICS Studies the rate at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).

4. OUTLINE: KINETICS Reaction Rates How we measure rates. Rate Laws How the rate depends on amounts of reactants. Integrated Rate Laws How to calc amount left or time to reach a given amount. Half-life How long it takes to react 50% of reactants. Arrhenius Equation How rate constant changes with T. Mechanisms Link between rate and molecular scale processes.

5. FACTORS THAT AFFECT REACTION RATES Concentration of Reactants ◦As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. Temperature ◦At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. Catalysts ◦Speed rxn by changing mechanism.

6. REACTION RATES Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. [A] vs t Rxn Movie

7. THE RATES OF REACTIONS Depends on composition and temperature of reaction mixture. (a) Definition of rate - as the slope of the tangent drawn to the curve showing the variation of concentration with time.

8. REACTION RATES In this reaction, the concentration of butyl chloride, C 4 H 9 Cl, was measured at various times, t. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) [C 4 H 9 Cl] M

9. REACTION RATES The average rate of the reaction over each interval is the change in concentration divided by the change in time: C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Average Rate, M/s [C 4 H 9 Cl]

10. REACTION RATES Note that the average rate decreases as the reaction proceeds. This is because as the reaction goes forward, there are fewer collisions between reactant molecules. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

11. REACTION RATES A plot of concentration vs. time for this reaction yields a curve like this. The slope of a line tangent to the curve at any point is the instantaneous rate at that time. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

12. REACTION RATES The reaction slows down with time because the concentration of the reactants decreases. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

13. REACTION RATES AND STOICHIOMETRY In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Rate = -[C 4 H 9 Cl] t = [C 4 H 9 OH] t

14. REACTION RATES AND STOICHIOMETRY What if the ratio is not 1:1? H 2 (g) + I 2 (g)  2 HI (g) Only 1/2 HI is made for each H 2 used.

15. REACTION RATES AND STOICHIOMETRY To generalize, for the reaction aA + bBcC + dD Reactants (decrease)Products (increase)

16. CONCENTRATION AND RATE Each reaction has its own equation that gives its rate as a function of reactant concentrations.  this is called its Rate Law To determine the rate law we measure the rate at different starting concentrations.

17. CONCENTRATION AND RATE Compare Experiments 1 and 2: when [NH 4 + ] doubles, the initial rate doubles.

18. CONCENTRATION AND RATE Likewise, compare Experiments 5 and 6: when [NO 2 - ] doubles, the initial rate doubles.

19. CONCENTRATION AND RATE This equation is called the rate law, and k is the rate constant.

20. The rate of change of molar conc of CH 3 radicals in the reaction 2CH 3 (g) CH 3 CH 3 (g) was reported as d[CH 3 ]/dt=-1.2 mol.dm -3 s -1 under particular conditions. What is (a) the rate of reaction (b) the rate of formation of CH 3 CH 3 ? EXAMPLE 1

21. (a) What is the rate of reaction ? 2CH 3 (g) CH 3 CH 3 (g) d[CH 3 ]/dt=-1.2 mol dm -3 s -1 Rate of this reaction = -(1/2) x (-1.2 mol dm -3 s -1 ) = 0.6 mol dm -3 s -1

22. (b) the rate of formation of CH 3 CH 3 ? 2CH 3 (g) CH 3 CH 3 (g) d[CH 3 ]/dt=-1.2 mol dm -3 s -1 Rate of this formation = -(1/2) x (-1.2 mol dm -3 s -1 ) = 0.6 mol dm -3 s -1

23. RATE LAWS A rate law shows the relationship between the reaction rate and the concentrations of reactants. ◦For gas-phase reactants use P A instead of [A]. k is a constant that has a specific value for each reaction. The value of k is determined experimentally. “Constant” is relative here- k is unique for each rxn k changes with T K = dm 3 mol -1 s -1

24. RATE LAWS Exponents tell the order of the reaction with respect to each reactant. This reaction is First-order in [NH 4 + ] First-order in [NO 2 − ] The overall reaction order can be found by adding the exponents on the reactants in the rate law.

25. APPLICATION OF RATE LAW -To predict the rate of reaction from the composition of mixture -Guide to the mechanism of the reaction- for any proposed mechanism- must be consistent with the observed rate law.

26. INTEGRATED RATE LAWS Consider a simple 1st order rxn: A  B How much A is left after time t? Integrate: Differential form:

27. INTEGRATED RATE LAWS The integrated form of first order rate law: Can be rearranged to give: [A] 0 is the initial concentration of A (t=0). [A] t is the concentration of A at some time, t, during the course of the reaction.

28. INTEGRATED RATE LAWS Manipulating this equation produces… …which is in the form y = mx + b

29. FIRST-ORDER PROCESSES If a reaction is first-order, a plot of ln [A] t vs. t will yield a straight line with a slope of -k. So, use graphs to determine rxn order.

30. FIRST-ORDER PROCESSES Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 NCCH 3 CN How do we know this is a first order rxn?

31. FIRST-ORDER PROCESSES This data was collected for this reaction at 198.9°C. CH 3 NCCH 3 CN Does rate=k[CH 3 NC] for all time intervals?

32. FIRST-ORDER PROCESSES When ln P is plotted as a function of time, a straight line results. ◦The process is first-order. ◦k is the negative slope: 5.1  10 -5 s -1.

33. EXAMPLE 2 In particular experiment, it was found that the concentration of N 2 O 5 in liquid bromine varied with time as follows: 2N 2 O 5 4NO 2 + O 2 t/s [N 2 O 5 ]/mol dm -3 00.2 4000.4 8000.6 16000.8 20001.0

34. SOLUTION:

35. SOLUTION: To confirm the first order reaction, plot the ln ([A]/[A] o ) against the time & expect the straight line If a straight line is obtained, its slope can be identified with k.

36. INTEGRATED RATE LAW － FIRST ORDER

37. SECOND-ORDER PROCESSES Similarly, integrating the rate law for a process that is second-order in reactant A: also in the form y = mx + b Rearrange, integrate:

38. SECOND-ORDER PROCESSES So if a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line with a slope of k. If a reaction is first-order, a plot of ln [A] t vs. t will yield a straight line with a slope of -k. First order:

39. DETERMINING RXN ORDER Time (s)[NO 2 ], M 0.00.01000 50.00.00787 100.00.00649 200.00.00481 300.00.00380 The decomposition of NO 2 at 300°C is described by the equation NO 2 (g) NO (g) + 1/2 O 2 (g) and yields these data:

40. Graphing ln [NO 2 ] vs. t yields: Time (s)[NO 2 ], Mln [NO 2 ] 0.00.01000-4.610 50.00.00787-4.845 100.00.00649-5.038 200.00.00481-5.337 300.00.00380-5.573 DETERMINING RXN ORDER The plot is not a straight line, so the process is not first- order in [A]. Does not fit:

41. SECOND-ORDER PROCESSES Time (s)[NO 2 ], M1/[NO 2 ] 0.00.01000100 50.00.00787127 100.00.00649154 200.00.00481208 300.00.00380263 A graph of 1/[NO 2 ] vs. t gives this plot. This is a straight line. Therefore, the process is second-order in [NO 2 ].

42. HALF-LIFE Half-life is defined as the time required for one- half of a reactant to react. Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] 0.

43. HALF-LIFE For a first-order process, set [A] t =0.5 [A] 0 in integrated rate equation: NOTE: For a first-order process, the half-life does not depend on [A] 0.

44. HALF-LIFE- 2ND ORDER For a second-order process, set [A] t =0.5 [A] 0 in 2nd order equation.

45. OUTLINE: KINETICS

46. TEMPERATURE AND RATE Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent.

47. THE COLLISION MODEL In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other.

48. THE COLLISION MODEL Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

49. ACTIVATION ENERGY In other words, there is a minimum amount of energy required for reaction: the activation energy, E a. Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

50. REACTION COORDINATE DIAGRAMS It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile.

51. Reaction Coordinate Diagrams It shows the energy of the reactants and products (and, therefore, E). The high point on the diagram is the transition state. The species present at the transition state is called the activated complex. The energy gap between the reactants and the activated complex is the activation energy barrier.

52. Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and E a : where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. Ea : Activation Energy

53. Arrhenius Equation Taking the natural logarithm of both sides, the equation becomes y = mx + b When k is determined experimentally at several temperatures, E a can be calculated from the slope of a plot of ln k vs. 1/T.

54. EXAMPLE 3 The rate of the second order decomposition of acetaldehyde (ethanal, CH 3 CHO) was measure over the temperature range 700-1000K, and the rate constant are reported as below. Find the E a and A. T/K7007307607908108409101000 kr/dm 3 mol -1 s -1 0.0110.0350.1050.3430.7892.1720145

55. SOLUTION: The data can be analysed by plotting ln(kr/dm 3 mol -1 s -1 ) against 1/(T/K) or more conveniently (10 3 K)/T (10 3 )K/T1.431.371.321.271.231.191.101.00 ln (kr/dm 3 mol -1 s -1 )-4.51-3.35-2.25-1.07-0.240.773.004.98 The least square fit is to a line with slope -22.7 and intercept 27.7.

56. Therefore: Ea= 22.7 x (8.314 JK -1 mol -1 ) x 103 K = 189 kJmol -1 A= e 27.7 dm 3 mol -1 s -1 = 1.1 x 10 12 dm 3 mol -1 s -1 y = mx + b

57. Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism. Reactions may occur all at once or through several discrete steps. Each of these processes is known as an elementary reaction or elementary process.

58. Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process. The rate law for an elementary step is written directly from that step.

59. Multistep Mechanisms In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate-determining step.

60. Slow Initial Step The rate law for this reaction is found experimentally to be Rate = k [NO 2 ] 2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps. NO 2 (g) + CO (g)  NO (g) + CO 2 (g)

61. Slow Initial Step A proposed mechanism for this reaction is Step 1: NO 2 + NO 2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast) The NO 3 intermediate is consumed in the second step. As CO is not involved in the slow, rate determining step, it does not appear in the rate law.

62. Fast Initial Step The rate law for this reaction is found (experimentally) to be Because termolecular (= trimolecular) processes are rare, this rate law suggests a two-step mechanism.

63. Fast Initial Step A proposed mechanism is Step 1 is an equilibrium- it includes the forward and reverse reactions.

64. Fast Initial Step The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be But how can we find [NOBr 2 ]?

65. Fast Initial Step NOBr 2 can react two ways: ◦With NO to form NOBr ◦By decomposition to reform NO and Br 2 The reactants and products of the first step are in equilibrium with each other. Therefore, Rate f = Rate r

66. Fast Initial Step Because Rate f = Rate r, k 1 [NO] [Br 2 ] = k −1 [NOBr 2 ] Solving for [NOBr 2 ] gives us k 1 k −1 [NO] [Br 2 ] = [NOBr 2 ]

67. Fast Initial Step Substituting this expression for [NOBr 2 ] in the rate law for the rate-determining step gives

68. Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs.

69. Catalysts One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.

70. Enzymes Enzymes are catalysts in biological systems. The substrate fits into the active site of the enzyme much like a key fits into a lock.

71. Exercise: The rate constant for the first order decomposition of a compound A In the reaction 2 N 2 (g)  O 2 (g) is kr=3.56 x 10 -7 s -1 at 25 o C. a)Calculate is the half A? b) Calculate what will be the pressure, initially 33.0 kPa at (i) 50s (ii) 20 min after the initiation of the reaction?