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Kinetics: Rates and Mechanisms of Reactions

Kinetics: Rates and Mechanisms of Reactions. Chemical Kinetics tells us: …how fast a reaction will occur …how molecules react (MECHANISM). a mechanism is a sequence of steps that lead to the product. Factors Affecting the rate:. 1. concentration:. 2. temperature:.

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Kinetics: Rates and Mechanisms of Reactions

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  1. Kinetics: Rates and Mechanisms of Reactions Chemical Kinetics tells us: …how fast a reaction will occur …how molecules react (MECHANISM) a mechanism is a sequence of steps that lead to the product

  2. Factors Affecting the rate: 1. concentration: 2. temperature: generally a 10oC increase will double the rate 3. nature of the reactant: i.e. surface area 4. catalyst: (two types: homogenous and heterogeneous) 5. mechanism: (orientation, shape, & order) COLLISION THEORY = CAPPING A MARKER

  3. Rate of RXN = The increase in concentration of a product per unit time. or The decrease in concentration of a reactant per unit time.

  4. Conc. is usually measured in M (Molarity= mol/L) for solutions. Rate = M time = mol L • s or mol•L-1•s-1 or M•s -1 Square brackets [ ] are often used to express molarity (i.e. [HCl] means Molarity of HCl) Since many reactions involve gases, P is often used for concentration. moles/L

  5. Consider the reaction (net ionic eq.): 3ClO- (aq)  2Cl-(aq) + ClO3-(aq) Rate could be defined in at least 3 ways: (3 coefficients and ions) 1. appearance of ClO3- 2. appearance of Cl- Disappearance 3. disappearance of ClO-

  6. Consider the reaction (net ionic): 3ClO- (aq)  2Cl-(aq) + ClO3-(aq) 1. appearance of ClO3- 2. appearance of Cl- Disappearance 3. disappearance of ClO- Question: Are these three rates equal?

  7. Consider the reaction (net ionic): 3ClO- (aq)  2Cl-(aq) + ClO3-(aq) Let’s make these three rates equal. Note the use of coefficients and the - sign

  8. General Form: aA + bB  cC + dD = -[B] btime = -[A] atime = [D] dtime rate =  [C] ctime “REACTANTS” “PRODUCTS”

  9. Average rate = slope (over time period) negative sign

  10. instantaneous rate = tangent slope (changing) WHY? Collision Theory!

  11. IF we can now somehow get a linear plot in the form of: y = mx + b. The slope would be a constant independent of concentration!

  12. We could call the slope the rate constant and assign it the letter k! rate constant = k  instantaneous rate or rate “call in the mathematicians”

  13. rate constant: k is conc. independent Is still temperature dependent and mechanism dependent! General form of rate law : for RXN: A  products m = RXN order according to A. Determined by experiment only! conc. of A rate constant

  14. General form of rate law : RXN orders (w, x, y, and z)must be determined by exp. only! Total (overall) order =  individual orders

  15. General Equation Forms: 0 order: rate = k 1st order: rate = k[A] 2nd order: rate = k[A]2 or rate = k[A][B] 3rd order: rate = k[A]3 or ........ Simple experiments are done by doubling 1 conc. at a time and looking at the effect.

  16. General Equation Forms: 0 order: rate = k 1st order: rate = k[A] 2nd order: rate = k[A]2 or rate = k[A][B] 3rd order: rate = k[A]3 or ........ Simple experiments are done by doubling 1 conc. at a time and looking at the effect. order doubling effect on rate 0 [2]0 = 1 none 1/2 [2]1/2 = 1.41.. increase by 1.41.. 1 [2]1 = 2 doubles 2 [2]2 = 4 quadruples Question: suppose rate = k[A]2[B] what is the effect of doubling both A and B?

  17. Let’s look a some rate data for the RXN: NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)

  18. NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l) doubles double doubles double

  19. NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l) 2 2 2 2 for NO2- : [2]y = 2 y = 1 (1st Order) for NH4+ : [2]x = 2 x = 1 (1st Order)

  20. first order with respect to each reactant, 2nd order overall. orders usually have integer values, but can be fractional. Can also be (-) (inhibitors). Since rate has units, k must also have units. so units of k must work with [ ] to match units.

  21. Determine the units of k in each of the following: Since rate has units, k must also have units. (so units of k must work with [ ] to match units.) 1. rate = k[A] M kunits must = time-1 ? 2. rate = k[A]2 M2 kunits must = time-1 M-1 ? so k must have units of M-1t-1

  22. 4. rate = k[A][B]2[C] kunits=M-3time-1 5. rate = k[A]0

  23. Ebbing 4th ed. P 490 14.38 In a kinetic study of the reaction: 2NO(g) + O2(g)  2NO2(g) the following data were obtained for the initial rates disap- pearance of NO: Obtain the rate law. What is the value of the rate constant? .

  24. Ebbing 4th ed. P 490 14.38 In a kinetic study of the reaction: 2NO(g) + O2(g)  2NO2(g) the following data were obtained for the initial rates disap- pearance of NO: Obtain the rate law. What is the value of the rate constant? Write overall rate equation: rate = k[NO]x[O2]y For NO: select a pair of experiments in which the conc. of [NO] is changed, but other concentrations are unchanged. Let’s use Exp. 1 and 2

  25. Ebbing 4th ed. P 490 14.38 In a kinetic study of the reaction: 2NO(g) + O2(g)  2NO(g) Obtain the rate law. What is the value of the rate constant? overall rate equation: rate = k[NO]x[O2]y “Let’s divide Exp.1 by Exp.2 to allow us to cancel terms.

  26. Ebbing 4th ed. P 490 14.38 In a kinetic study of the reaction: 2NO(g) + O2(g)  2NO(g) Obtain the rate law. What is the value of the rate constant? overall rate equation: rate = k[NO]x[O2]y

  27. Ebbing 4th ed. P 490 14.38 In a kinetic study of the reaction: 2NO(g) + O2(g)  2NO(g) overall rate equation: rate = k[NO]x[O2]y How do we solve for x? Use logarithms

  28. Ebbing 4th ed. P 490 14.38 In a kinetic study of the reaction: 2NO(g) + O2(g)  2NO(g) overall rate equation: rate = k[NO]x[O2]y How do we solve for x? Use logarithms Therefore the rate equation becomes: rate = k[NO]2[O2]y

  29. Now let’s determine the reaction order with respect to [O2]

  30. Ebbing 4th ed. P 490 14.38 In a kinetic study of the reaction: 2NO(g) + O2(g)  2NO2(g) the following data were obtained for the initial rates disap- pearance of NO: Write overall rate equation: rate = k[NO]2[O2]y For O2: select a pair of experiments in which the conc. of [O2] is changed, but all other concentrations are unchanged. Let’s use Exp. 1 and 3

  31. Ebbing 4th ed. P 490 14.38 In a kinetic study of the reaction: 2NO(g) + O2(g)  2NO2(g) the following data were obtained for the initial rates disap- pearance of NO: Write overall rate equation: rate = k[NO]2[O2]y Let’s use Exp. 1 and 3

  32. Ebbing 4th ed. P 490 14.38 In a kinetic study of the reaction: 2NO(g) + O2(g)  2NO2(g) Write overall rate equation: rate = k[NO]2[O2]y

  33. Ebbing 4th ed. P 490 14.38 In a kinetic study of the reaction: 2NO(g) + O2(g)  2NO2(g) Write overall rate equation: rate = k[NO]2[O2]y y = 1 rate = k[NO]2[O2]1 Therefore: Now solve for k.

  34. Ebbing 4th ed. P 490 14.38 In a kinetic study of the reaction: 2NO(g) + O2(g)  2NO2(g) Overall rate equation: rate = k[NO]2[O2]1 Choose any Exp. and substitute in experimental to obtain k. i.e. Exp1. : rate = k[NO]2[O2]1 so : 0.0281 = k[0.0125]2[0.0253] k = 7100 M-2s-1

  35. Ebbing 4th ed. P 490 14.40 Iodine ion is oxidized to hypoiodite ion, IO-, by hypochlorite ion, ClO-, in basic solution. the following initial rate experiments were run: Obtain the rate law. What is the value of the rate constant? .

  36. k = 6.1 s-1

  37. This data isn’t linear! What can we do? Integrated Rate Laws

  38. IF we can now somehow get a linear plot in the form of: y = mx + b. The slope would be a constant, independent of concentration!

  39. We could call the slope the rate constant and assign it the letter k! rate constant = k  rate “call in the mathematicians”

  40. Key Equations: *Since the units of rate are concentration/time, the units of k (the rate constant) must dimensionally agree. So for each order, k will have different units and those units can tell one which equation to use. [ ] means the concentration of the enclosed species in Molarity (M).

  41. The data below was collected for the reaction: NOCl(g)  NO(g) + 1/2Cl2(g) Time (s) [NOCl] (M) 0 0.100 30 0.064 60 0.047 100 0.035 200 0.021 300 0.015 400 0.012 Prepare THREE graphs to determine if the RXN is ZERO, 1st, or 2nd order. Then determine the value and units of the rate constant k.

  42. Zero Order Plot [A]t vs. time rate = k [A]t = -kt + [A]0 y = mx + b

  43. First Order Plot ln[A]t vs. time rate =k[A] ln[A]t = -kt + ln[A]0 y = mx + b

  44. 2nd Order Plot 1/[A]t vs. time rate = k[A]2 1/[A]t = kt + 1/[A]0 y = mx + b Plot is linear so 2nd Order k = slope = 0.185 M-1s-1

  45. Integrated rate laws: Zero Order: rate = k[A]0 = k rate = k integrated gives: [A]t = -kt + [A]0 y = mx + b slope = -k If a RXN is zero order, a plot of [A] vs. time should be linear and the slope = -k.

  46. Integrated rate laws: 1st order rate laws: rate = k[A] integrated gives: rearranged to : y = mx + b gives: ln[A]t = -kt + ln[A]0 slope = -k If reaction data is 1st order, a plot of ln[A] vs. time should be linear.

  47. 2nd Order Integrated Rate Equations: integrated gives: rate = k[A]2 y = mx + b slope = k If a RXN is 2nd order, a plot of 1/[A] vs. time should be linear and the slope = k.

  48. RXN is first order with respect to CH3NC Zero Order Plot 1st Order Plot slope = -k ln[CH3NC]t = -kt + ln[CH3NC]0

  49. RXN is 2nd order with respect to [NO2] First order plot 2nd order Plot Slope = k

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