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Chapter 15 - Chemical Kinetics

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  1. Chapter 15 - Chemical Kinetics • Objectives: • Determine rates of reactions from graphs of concentration vs. time. • Recall the conditions which affect the rates. • Recognize the order of reaction, give the rate equation, calculate the rate constant. • Use integrated rate laws and half-life equations in calculations. • Draw energy diagrams, find activation energy. • Identify catalysts and their properties. • Identify reaction intermediates. • Recognize the rate equation given a mechanism, and given the rate equation determine the mechanism.

  2. KINETICS — the study of REACTION RATES H2O2 decomposition in an insect H2O2 decomposition catalyzed by MnO2 Chemical engineering, enzymology, environmental engineering, etc.

  3. Kinetics and Mechanisms • KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM. • The reaction mechanism is our goal! • The sequence of events at the molecular level that control the speed and outcome of a reaction.

  4. Br from biomass burning destroys stratospheric ozone. (See R.J. Cicerone, Science, volume 263, page 1243, 1994.) Step 1: Br + O3 ---> BrO + O2 Step 2: Cl + O3 ---> ClO + O2 Step 3: BrO + ClO + light ---> Br + Cl + O2 NET: 2 O3 ---> 3 O2 • Identify the intermediate • Identify the catalyst(s)

  5. D concentration D time Rate = Change in distance over a period of time: D distance D time Reaction Rates • Reaction rate = change in concentration of a reactant or product with time. • Three “types” of rates: • initial rate • average rate • instantaneous rate

  6. Reaction Rates can be determined from a Plot • The initial rate (in this case over the first minute) is calculated from a tangent line crossing the initial concentration. Then the slope of the line is determined: • Blue dye is oxidized with bleach. • Its concentration decreases with time. • The rate — the change in dye conc with time — can be determined from the plot.

  7. Reaction Rates can be determined from a Plot 2 N2O5 NO2 + O2 4 • The average rate is calculated from a time interval. • The instantaneous rate is calculated at a single point in time (or given concentration) by drawing a tangent line crossing the point. Then the slope of the line is determined. • Compare average rates at the beginning and end of reaction.

  8. Factors affecting the Rates • Concentrations • Physical State of Reactants and Products • Surface area • Temperature • Catalysts Review Exp. 1: Factors affecting reactions rates

  9. Concentration Mg(s) + 2 HCl(aq) ---> MgCl2(aq) + H2(g) • Increasing the concentration of reactants _______________ the rate of the reaction. 0.3 M HCl 6 M HCl

  10. Surface Area • Increasing the surface area of reactants _____________ the reaction rate.

  11. Temperature • Increasing the temperature ____________ the rate of the reaction. Bleach at 54 ˚C Bleach at 22 ˚C

  12. Catalysts • A _________ is present at the beginning and at the end of the reaction and it does not change; but it _______________ the rate of the reaction. MnO2 2 H2O2 ---- > 2 H2O + O2

  13. Factors affecting Reaction Rates Iodine clock reaction: 1. Iodide is oxidized to iodine H2O2 + 2 I- + 2 H+ -----> 2 H2O + I2 2. I2 reduced to I- with vitamin C I2 + C6H8O6 ----> C6H6O6 + 2 H+ + 2 I- When all vitamin C is depleted, the I2 interacts with starch to give a blue complex.

  14. Factors affecting Reaction Rates

  15. Concentration and Rate • To postulate a mechanism we study: • - The reaction rate • and its • - Concentration dependence. • Generate a Rate Law Equation.

  16. Rate Laws • In general for: a A + b B --> x X Rate = k [A]m[B]n The exponents m, n • are the _______________ • can be 0, 1, 2 or fractions • must be determined by ____________! With a catalyst C [C]p

  17. Interpreting Rate Laws Rate = k [A]m[B]n[C]p • If m = 1, rxn. is 1st order in A Rate = k [A]1 If [A] doubles, then rate goes up by factor of __________ • If m = 2, rxn. is 2nd order in A. Rate = k [A]2 Doubling [A] increases rate by _____________ • If m = 0, rxn. is zero order. Rate = k [A]0 If [A] doubles, rate ____________________

  18. Deriving Rate Laws Derive rate law and k for CH3CHO(g) --> CH4(g) + CO(g) from experimental data for rate of disappearance of CH3CHO Expt. [CH3CHO] Disappear of CH3CHO (mol/L) (mol/L•sec) 1 0.10 0.020 2 0.20 0.081 3 0.30 0.182 4 0.40 0.318 Rate = k [CH3CHO]n 4 Rate = k [2 CH3CHO]n n = 2 Look at exp 1 and 2: concentration doubles; rate cuadruples

  19. Deriving Rate Laws Rate of rxn = Here the rate goes up by ________ when initial concentration doubles. Therefore, we say this reaction is __________ order. Now determine the value of k. Use any exp. Data: Using k you can calculate rate at other values of [CH3CHO] at same T.

  20. Concentration and Time • What is the concentration of reactant as a function of time? • Consider First Order reactions: Integrating we get:

  21. Integrated First Order Law [A] / [A]0 =fraction remaining after time t has elapsed.

  22. The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 year-1 at 12oC. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 x 10-7 g/cm3. Assume that the average temperature of the lake is 12oC. • a) What is the concentration of the insecticide on June 1 of the following year?b) How long will it take for the concentration of the insecticide to drop to 3.0 x 10-7 g/cm3?

  23. Using Integrated Rate Laws All 1st order reactions have straight line plot for ln [A] vs. time. And 2nd order gives straight line for plot of 1/[A] vs. time.

  24. Using Integrated Rate Laws • In an experiment for: 2 N2O5(g) ---> 4 NO2(g) + O2(g) Time (min) [N2O5]0 (M) 0 1.00 1.0 0.705 2.0 0.497 5.0 0.173 If it were zero order: Data of conc. vs. time plot do not fit straight line.

  25. Using Integrated Rate Laws • In an experiment for: 2 N2O5(g) ---> 4 NO2(g) + O2(g) Time (min) [N2O5]0 (M) 0 1.00 1.0 0.705 2.0 0.497 5.0 0.173 If it were first order: ln [N2O5]0 0 -0.35 -0.70 -1.75 Calculate the ln : Plot of ln [N2O5] vs. time is a straight line!

  26. Using Integrated Rate Laws Plot of ln [N2O5] vs. time is a straight line! Eqn. for straight line: y = mx + b

  27. The gas phase decomposition of hydrogen peroxide at 400 oC is second order in H2O2. In one experiment, when the initial concentration of H2O2 was 0.246 M, the concentration of H2O2 dropped to 3.39 x 10-2 M after 25.9 seconds had passed. What is the rate constant for the reaction? 2 H2O2 2 H2O + O2

  28. Half-Life HALF-LIFE is the time it takes for 1/2 a sample is disappear. For 1st order reactions, the concept of HALF-LIFE is especially useful.

  29. Half-Life • Reaction is 1st order decomposition of H2O2. • Reaction after 1 half-life. • 1/2 of the reactant has been consumed and 1/2 remains.

  30. Half-Life • After 2 half-lives _____ of the reactant remains. • After 3 half-lives ____ of the reactant remains.

  31. Half-Life ln [R] / [R]0 = k t [A] / [A]0 = fraction remaining when t = t1/2 then fraction remaining = _________ ln (____) = - k • t1/2

  32. In an experiment, it is determined that 75% of a sample of HCO2H (formic acid) has decomposed in 72 seconds following first-order kinetics. Determine t1/2 for this reaction. HCO2H  CO2 + H2 ln [R] / [R]0 = k t t1/2 = 0.693 / k

  33. Mechanisms • Mechanism: how reactants are converted to products at the molecular level. RATE LAW ----> MECHANISM experiment ----> theory

  34. Activation Energy Molecules need a minimum amount of energy to react. Visualized as an energy barrier - activation energy, Ea. Reaction coordinate diagram

  35. Activation Energy • Conversion of cis to trans-2-butene requires twisting around the C=C bond. • Rate = k [trans-2-butene]

  36. Activation energy barrier Transition State Cis Transition state Trans

  37. Mechanisms • Reaction passes thru a TRANSITION STATEwhere there is an activated complex that has sufficient energy to become a product. ACTIVATION ENERGY, Ea = energy req’d to form activated complex. Here Ea = ___________

  38. Mechanisms Also note that trans-butene is MORE STABLE than cis-butene by about 4 kJ/mol. Therefore, cis ---> trans is _________________ This is the connection between thermo-dynamics and kinetics.

  39. Effect of Temperature In ice at 0 oC • Reactions generally occur slower at lower T. Room temperature Iodine clock reaction, book page 705. H2O2 + 2 I- + 2 H+ --> 2 H2O + I2

  40. Activation Energy and Temperature Reactions are __________ at a higher T because a larger fraction of reactant molecules have enough energy to convert to product molecules. In general, differences in activation energy cause reactions to vary from fast to slow.

  41. Collision Theory • Molecules must collide with one another • Molecules must collide with sufficient ________ to break bonds • Molecules must collide in an orientation that can lead to rearrangement of the atoms.

  42. Temp (K) Rate constant 8.31 x 10-3 kJ/K•mol Activation energy Frequency factor Arrhenius Equation Frequency factor related to frequency of collisions with correct geometry. Plot ln k vs. 1/T ---> straight line. slope = -Ea / R

  43. Collision Theory Reactions require (a) activation energy and (b) correct geometry. O3(g) + NO(g) ---> O2(g) + NO2(g) 2. Activation energy and geometry 1. Activation energy

  44. Mechanisms Most reactions involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] NOTE 1. Rate law comes from experiment. 2. Order and stoichiometric coefficients not necessarily the same! 3. Rate law reflects all chemistry down to and including the slowest step in multistep reaction.

  45. Mechanisms Most reactions involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] Proposed Mechanism: Step 1 — slow HOOH + I- --> HOI + OH- Step 2 — fast HOI + I- --> I2 + OH- Step 3 — fast 2 OH- + 2 H+ --> 2 H2O Rate of the reaction controlled by slow step — RATE DETERMINING STEP, Rate can be no faster than RDS!

  46. Mechanisms Most reactions involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] Elementary Step 1 is bimolecular and involves I- and HOOH. Therefore, this predicts the rate law should be Rate a [I-] [H2O2] — as observed!! The species HOI and OH- are ________________. Proposed Mechanism: Step 1 — slow HOOH + I- --> HOI + OH- Step 2 — fast HOI + I- --> I2 + OH- Step 3 — fast 2 OH- + 2 H+ --> 2 H2O

  47. Uncatalyzed reaction Catalyzed reaction Catalysts and Activation Energy MnO2 catalyzes decomposition of H2O2 2 H2O2 ---> 2 H2O + O2

  48. Catalysts and Activation Energy • Iodine-Catalyzed Isomerization of Cis-2-Butene

  49. Remember • Go over all the contents of your textbook. • Practice with examples and with problems at the end of the chapter. • Practice with OWL tutor. • Practice with the quiz on CD of Chemistry Now. • Work on your OWL assignment for Chapter 15.