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Rates of Reaction: Understanding Chemical Kinetics

Explore the concept of reaction rates and the factors that affect them. Learn about rate equations, concentration-time relationships, and activation energy. Gain a deeper understanding of chemical kinetics.

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Rates of Reaction: Understanding Chemical Kinetics

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  1. Chapter 13 Rates of Reaction Suroviec Spring 2014

  2. I. Reaction Rates • Chemical Kinetics: the investigation of the rate at which reactions occur • Rate: how fast a quantity changes with time

  3. I. Reaction Rates Rate = change in quantity / time elapsed

  4. A. Average Rate Vs. Instantaneous Rate

  5. B. Writing Rate Reactions • Consider the rate reaction: 2NO2 (g)  2NO (g) + O2 (g)

  6. General form aA + bB  cC + dD

  7. Example • Give the relative rates of disappearance of the reactants and formation of products for each of the following reactions: • 2O3 (g)  3O2 (g) • 2HOF (g)  2HF (g) + O2 (g)

  8. Example • A  2B • Rate increases or decreases as there is less A to react • How is rate of change of [A] related to rate of change of [B]? • Find rate of change of [A] for time internal from 10.0 to 20.0 s • What is the instantaneous rate when [B] = 0.750M

  9. Example N2 (g) + 3H2 (g)  2NH3 (g) What are the relative rates? If –D[H2]/Dt = 4.5X10-4 M/min, What is –D[N2]/Dt? What is D[NH3]/Dt?

  10. II. Rate Laws • Why look at initial rates? • Ex. 2N2O5 4NO2 + O2

  11. General Form of the Rate Law aA + bB  cC + dD Rate = k[A]x[B]y

  12. A. Method of Initial Rates • Initial rate of reaction is measured with several different starting amount of reactants

  13. Example The reaction between ozone and nitrogen dioxide at 231 K is first order in both [NO2] and [O3]. 2 NO2 (g)+ O3 (g)  N2O5 (g) + O2(g) • Write the rate equation for the reaction • If the concentration of NO3 is tripled, what is the change in the reaction rates? • What is the effect on the reaction rate if the concentration of the O3 is halved?

  14. Example 2NO(g) + O2 (g) 2NO2 (g) • Determine the order of the reaction for each reactant • Write the rate equation for the reaction • Calculate the rate constant • Calculate the rate in M/s at the instant when [NO] = 0.015 M and [O2] = 0.0050M

  15. Example A + B  C What is the rate equation?

  16. Base 10 log and Natural Log • Base 10 log: 102 = 100  log 100 = _______ 103 = 1000  log 1000 = _______ log 45 = 1.65  101.65 =_________ log x = y  10y =x • Natural Log: base is e = 2.718281828….. e1 = 2.71828 ln(2.71828) = 1 ln(e)=_______ ln(45) = 3.81  e3.81 =_________ ln x = y  ey =x

  17. Important properties of both log and ln • log ab = log a + log b • log a/b = log a – log b • log ab = b(log a)

  18. Example • Rate = k[A]2[B]x

  19. III. Concentration-time relationships • Useful to know • How long a reaction is going to proceed to reach concentration of interest for product or reactant • After a given time what are the concentrations of the reactants and the products?

  20. A. 1st order reactions • A  B • Rate = k[A] • Integrated rate eqn for 1st order reactions: • Useful for 3 reasons • Can calculate k if we know [A]/[A]o • if k and [A]o are known then we can determine the amount of material expected after time t ([A]) • If k is known than after time t we can calculate the fraction [A]/[A]o

  21. B. 2nd order reactions • Rate = k[A]2 • Integrated rate eqn for 2nd order reactions: 1/[A] = kt + 1/[A]o OR 1/[A] – 1[A]o = kt

  22. Example Ammonium cyanate, NH4NCO, rearranges in water to give urea, (NH2)2CO. NH4NCO (aq)  (NH2)2CO (aq) The rate equation for this process is: Rate = k [NH4NCO]2 where k = 0.0113 L/mol.min. If the original concentration of NH4NCO in solution is 0.229M how long will it take for the concentration of decrease to 0.180M?

  23. Example Sucrose breaks down to form glucose and fructose. C12H22O11(aq) +H2O (l)  2C6H12O6 (aq) • Plot ln sucrose versus time • and 1/[sucrose] versus time. • Find the order of the reaction • Write the rate constant for the • reaction and calculate k • Find concentration of sucrose • After 175 min

  24. C. Zero Order Reactions • Rate =k[A]0 • Integrated rate eqn for 0th order reactions: [A]o – [A] = kt

  25. Table 15-1, p.719

  26. D. Half-Life • Time required for half of a reactant to be consumed 1. 1st order

  27. Fig. 15-9, p.720

  28. Example The rate equation for the decomposition of producing NO2 and O2 is –D[N2O5]/Dt = k[N2O5] The value of k = 5.0 X 10-4 s-1 for the reaction at a known temperature. • Calculate the half-life of N2O5 • How long does it take for the N2O5 concentration to drop to 1/10th of its original value?

  29. 2. 2nd order and 0th order • A  B • Rate = k[A]2 • A  B • Rate = k

  30. IV. Activation Energy NO2 (g) + F2 (g)  FNO2 (g) + F (g) What makes this reaction occur? What makes this reaction not occur?

  31. A. Collision Theory • Reactant Molecules must collide with each other • Reactant molecules must collide with sufficient energy • Reactant molecules must have proper orientation 2 questions: • What makes a reaction go faster? • What does sufficient energy mean?

  32. A. What makes a reaction go faster? 1. 2.

  33. B. What does sufficient energy mean? RXN: AB + C  A + BC Mechanism:

  34. C. How can we measure Ea? • From collision theory Rate of reaction = (collision freq)X(fraction with proper orientation)X (fraction of collisions with enough energy) k = Ae-Ea/RT Arrhenuis equation

  35. Example • When heated to a high temperature, cyclobutane, C4H8, decomposes to ethylene. C4H8 (g)  2C2H4 (g) The activation energy, Ea for this reaction is 260 kJ/mol. At 800K the ate constant k = 0.0315 s-1. Determine the value of k at 850K.

  36. Uncatalyzed reaction Catalyzed reaction D. What is a catalyst? • A substance added to a reaction to speed it up, but not consumed in the reaction MnO2 catalyzes decomposition of H2O2 2 H2O2 ---> 2 H2O + O2

  37. V. Reaction mechanisms AB + C  A + BC Possible mechanisms

  38. V. Reaction Mechanisms • Molecularity • Elementary step

  39. A. Mechanisms and Rate Laws • For net reactions, no simple relationship can be assumed between reaction coefficients and reactant orders 2. For each elementary step reactant coefficients DO equal reactant orders

  40. A. Mechanisms and Rate Laws • The rate of a net reaction is essentially that of the slowest step (the rate determining step)

  41. A. Mechanisms and Rate Laws 4. When a fast step proceeds a slow step the fast step can be assumed to be at equilibrium RXN: H2 (g) + CO (g)  H2CO (g) Step 1. H2 (g)  2 H (g) fast Step 2. H (g) + CO (g)  HCO (g) slow Step 3. H (g)  H2CO (g) fast

  42. A. Mechanisms and Rate Laws 5. A catalyst can provide a new mechanistic pathway for a reaction 2Ce4+ (aq) + Tl+ (aq)  2Ce3+ (aq) + Tl3+ (aq) Mechanism Rate equation Add Mn2+ catalyst

  43. In General • Measure reaction rates • Through initial rates • Graphically concentraion vs. time • Formulate the rate law • Postulate the mechanism • Cannot prove, can only disprove

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