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Chemical Kinetics Chapter 12

Chemical Kinetics Chapter 12. H 2 O 2 decomposition in an insect. H 2 O 2 decomposition catalyzed by MnO 2. Objectives. Understand rates of reaction and the conditions affecting rates. Derive a rate of reaction, rate constant, and reaction order from experimental data

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Chemical Kinetics Chapter 12

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  1. Chemical KineticsChapter 12 H2O2 decomposition in an insect H2O2 decomposition catalyzed by MnO2

  2. Objectives • Understand rates of reaction and the conditions affecting rates. • Derive a rate of reaction, rate constant, and reaction order from experimental data • Use integrated rate laws • Discuss collision theory and the role of activation energy in a reaction • Discuss reaction mechanisms and their effect on rate law

  3. Chemical Kinetics • We can use thermodynamics to tell if a reaction is product- or reactant-favored. • But this gives us no info on HOW FAST reaction goes from reactants to products. • KINETICS— the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM.

  4. Reaction Mechanisms • A reaction mechanism is a sequence of events at the molecular level that control the speed and outcome of a reaction. • The reaction mechanism is our goal!

  5. Reaction Rates Section 12.1 • Reaction rate = change in concentration of a reactant or product with time. • Three “types” of rates • initial rate • average rate • instantaneous rate

  6. Factors Affecting Reaction Rates • Physical State of the Reactants • Gas, liquid or solid – how molecules are able to interact with each other • Solids react faster when surface area is greater so fine powders react faster than big chunks • 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 up reaction by changing mechanism. • Catalysts don’t get used up themselves

  7. Concentrations & Rates Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) 0.3 M HCl 6 M HCl

  8. Factors Affecting Rates • Physical state of reactants

  9. Factors Affecting Rates Catalysts: catalyzed decomp of H2O2 2 H2O2 2 H2O + O2

  10. Factors Affecting Rates • Temperature Bleach at 54 ˚C Bleach at 22 ˚C

  11. Determining a Reaction Rate 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. Dye Conc Time

  12. Determining a Reaction Rate

  13. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) Average Rate, M/s The average rate of the reaction over each interval is the change in concentration divided by the change in time:

  14. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • 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.

  15. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • 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.

  16. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • The reaction slows down with time because the concentration of the reactants decreases.

  17. Rxn Rate = -[C4H9Cl] t = [C4H9OH] t Reaction Rates and Stoichiometry C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1. • Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.

  18. Reaction Rates 2NO2(g) 2NO(g)+ O2(g) [C4H9Cl] M In this reaction, the concentration of nitrogen dioxide, NO2, was measured at various times, t.

  19. Reaction Rates and Stoichiometry 2NO2(g)  2NO(g) + O2(g) • What if the ratio is not 1:1? • 2NO can be made from 2NO2 consumed, but only 1 O2 is produced. • Read as: the rate of consumption of NO2 is the same as the rate of production of NO. This is because their coefficients are the same.

  20. Reaction Rates But since the coefficient for oxygen is ½ of the other two, it’s rate is of production rate is half as fast. Or 2 x rate O2 = rate of NO

  21. Reaction Rates • Reaction Rate and Stoichiometry summary: • For the reaction • C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) • we know • In general for • aA + bBcC + dD • Rxn

  22. Reaction Rates Practice • What is the relative rate of disappearance of the reactants and relative rate of appearance of the products for each reaction? • 2O3(g)  3O2(g) • 2HOF(g)  2HF(g) + O2(g) • In the synthesis of ammonia, if Δ[H2]/Δt =-4.5 x 10-4mol/L•min, what is rate with respect to NH3? wrt N2? N2(g) + 3H2(g)  2NH3(g)

  23. Concentrations & Rates 2NO2(g)  2NO(g) + O2(g) Rate of reaction is proportional to [NO2] We express this as a RATE LAW Rate of reaction = k [NO2]n where k = rate constant k is independent of conc. but increases with T n is the order of the reactant

  24. Concentrations, Rates, & Rate Laws In general, for a A + b B x X with a catalyst C Rate = k [A]m[B]n[C]p The exponents m, n, and p: • sum of m, n and p are the reaction order • can be 0, 1, 2 or fractions like 3/2 • must be determined by experiment! They are not simply related to stoichiometry!

  25. Ch. 12.3 - Interpreting Rate Laws Determining Rate by inspection (quick method): 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 _2_ • If m = 2, rxn. is 2nd order in A. Rate = k [A]2 Doubling [A] increases rate by ___4_____ • If m = 0, rxn. is zero order. Rate = k [A]0 If [A] doubles, rate _Stays the same_

  26. 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 just after the reactants are mixed. • Eliminates the effect of the reverse reaction.

  27. 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

  28. Deriving Rate Laws Determination of Rate by inspection: Here the rate goes up by __4___ when initial conc. doubles. Therefore, we say this reaction is ________2nd_________ order. So, the Rate of rxn = k [CH3CHO]2 Now determine the value of k. Use expt. #3 data— 0.182 mol/L•s = k (0.30 mol/L)2 k = 2.0 (L / mol•s) Using k you can calc. rate at other values of [CH3CHO] at same T.

  29. Deriving Rate Laws Determination of Rate by rigorous math: Rate = k[A]n Note: For easier math, put the faster rate in the numerator. 4=2n n=2

  30. Determination of Rate Law Practice For the reaction BrO3- + 5Br- + 6H+ 3Br2 + 3H2O • The general form of the Rate Law is Rate = k[BrO3-]n[Br-]m[H+]p • We use experimental data to determine the values of n, m, and p • We will choose trials that vary only one of the concentrations to determine the order based upon that reactant.

  31. Determination of Rate Law Practice Initial concentrations (M) BrO3- • Determine the rate law wrt each reactant • Determine the rate constant, k (with units) • Determine the overall order of reaction Br- H+ Rate (M/s) 0.10 0.10 0.10 0.8 x 10-3 0.20 0.10 0.10 1.6 x 10-3 0.20 0.20 0.10 3.2 x 10-3 0.10 0.10 0.20 3.2 x 10-3

  32. Ch. 12.4 – Integrated Rate Law (Concvs time) What is concentration of reactant as function of time? Consider FIRST ORDER REACTIONS The rate law is And the units of k are sec-1.

  33. 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). • Changes Rate = D[A]nDt • We will only work with n=0, 1, and 2

  34. Concentration/Time Relations Integrating - (∆ [A] / ∆ time) = k [A], we get ln is natural logarithm [A]0at time = 0 [A] / [A]0 =fraction remaining after time t has elapsed. Called the integrated first-order rate law.

  35. Concentration/Time Relations Sucrose decomposes to simpler sugars Rate of disappearance of sucrose = k [sucrose] If k = 0.21 hr-1 and [sucrose] = 0.010 M How long to drop 90% (to 0.0010 M)? Glucose

  36. Concentration/Time RelationsRate of disappear of sucrose = k [sucrose], k = 0.21 hr-1.If initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M? Use the first order integrated rate law ln (0.100) = - 2.3 = - (0.21 hr-1)(time) time = 11 hours

  37. Using the Integrated Rate Law The integrated rate law suggests a way to tell the order based on experiment. 2 N2O5(g)  4 NO2(g) + O2(g) Time (min) [N2O5] (M) ln [N2O5] 0 1.00 0 1.0 0.705 -0.35 2.0 0.497 -0.70 5.0 0.173 -1.75 Rate = k [N2O5]

  38. Using the Integrated Rate Law 2 N2O5(g)  4 NO2(g) + O2(g) Rate = k [N2O5] Plot of ln [N2O5] vs. time is a straight line! Data of conc. vs. time plot do not fit straight line.

  39. Using the Integrated Rate Law Plot of ln [N2O5] vs. time is a straight line! Eqn. for straight line: y = mx + b All 1st order reactions have straight line plot for ln [A] vs. time. ln[N2O5 ] = -kt + ln[N2O5]0    conc at rate constconc at time t =-slope time=0

  40. 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. See Active Figure 15.9

  41. Half-Life • Reaction is 1st order decomposition of H2O2.

  42. Half-Life • Reaction after 1 half-life. • 1/2 of the reactant has been consumed and 1/2 remains.

  43. Half-Life • After 2 half-lives 1/4 of the reactant remains.

  44. Half-Life • A 3 half-lives 1/8 of the reactant remains.

  45. Half-Life • After 4 half-lives 1/16 of the reactant remains.

  46. Half-Life Sugar is fermented in a 1st order process (using an enzyme as a catalyst). sugar + enzyme  products Rate of disappear of sugar = k[sugar] k = 3.3 x 10-4 sec-1 What is the half-life of this reaction?

  47. Half-Life Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the half-life of this reaction? Solution [A] / [A]0 = fraction remaining when t = t1/2 then fraction remaining = _______ Therefore, ln (1/2) = - k · t1/2 - 0.693 = - k · t1/2 t1/2 = 0.693 / k So, for sugar, t1/2 = 0.693 / k = 2100 sec = 35 min NOTE: For a first-order process, the half-life does not depend on [A]0.

  48. Half-Life Rate = k[sugar] and k = 3.3 x 10-4 sec-1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (140 min)? Solution 2 hr and 20 min = 4 half-lives Half-life Time Elapsed Mass Left 1st 35 min 2.50 g 2nd 70 1.25 g 3rd 105 0.625 g 4th 140 0.313 g

  49. Second Order • Rate = -Δ[A]/Δt = k[A]2 • integrated rate law • 1/[A] = kt + 1/[A]0 • y= 1/[A] m = k • x= t b = 1/[A]0 • A straight line if 1/[A] vs t is plotted • Knowing k and [A]0 you can calculate [A] at any time t

  50. NO2(g) NO (g) + 1/2 O2(g) Determining rxn order The decomposition of NO2 at 300°C is described by the equation and yields these data:

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