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Chemical Kinetics

Chemical Kinetics. Chapter 11. A B. rate =. D [A]. D [B]. rate = -. D t. D t. Chemical Kinetics. Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed?.

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Chemical Kinetics

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  1. Chemical Kinetics Chapter 11

  2. A B rate = D[A] D[B] rate = - Dt Dt Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). D[A] = change in concentration of A over time period Dt D[B] = change in concentration of B over time period Dt Because [A] decreases with time, D[A] is negative.

  3. A B rate = D[A] D[B] rate = - Dt Dt

  4. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) time 393 nm Detector light red-brown t1< t2 < t3 D[Br2] aD Absorption

  5. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) slope of tangent slope of tangent slope of tangent [Br2]final – [Br2]initial D[Br2] average rate = - = - Dt tfinal - tinitial instantaneous rate = rate for specific instance in time

  6. rate k = [Br2] rate a [Br2] rate = k [Br2] = rate constant = 3.50 x 10-3 s-1

  7. 2H2O2 (aq) 2H2O (l) + O2 (g) [O2] = P n V 1 1 D[O2] P = RT = [O2]RT RT RT DP rate = = Dt Dt measure DP over time PV = nRT

  8. 2A B aA + bB cC + dD rate = - = = rate = - = - D[C] D[B] D[A] D[B] D[D] D[A] rate = 1 1 1 1 1 Dt Dt Dt Dt Dt Dt c d a 2 b Reaction Rates and Stoichiometry Two moles of A disappear for each mole of B that is formed.

  9. D[CO2] = Dt D[CH4] rate = - Dt D[H2O] = Dt D[O2] = - 1 1 Dt 2 2 Write the rate expression for the following reaction: CH4(g) + 2O2(g) CO2(g) + 2H2O (g)

  10. aA + bB cC + dD The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. Rate = k [A]x[B]y Reaction is xth orderwith respect to A Reaction is yth orderwith respect to B Reaction is (x +y)th order overall

  11. F2(g) + 2ClO2(g) 2FClO2(g) rate = k [F2]x[ClO2]y Double [F2] with [ClO2] constant Rate doubles x = 1 rate = k [F2][ClO2] Quadruple [ClO2] with [F2] constant Rate quadruples y = 1

  12. F2(g) + 2ClO2(g) 2FClO2(g) 1 Rate Laws • Rate laws are always determined experimentally. • Reaction order is always defined in terms of reactant (not product) concentrations. • The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. rate = k [F2][ClO2]

  13. Determine the rate law and calculate the rate constant for the following reaction from the following data: S2O82-(aq) + 3I-(aq) 2SO42-(aq) + I3-(aq) rate k = 2.2 x 10-4 M/s = [S2O82-][I-] (0.08 M)(0.034 M) rate = k [S2O82-]x[I-]y y = 1 x = 1 rate = k [S2O82-][I-] Double [I-], rate doubles (experiment 1 & 2) Double [S2O82-], rate doubles (experiment 2 & 3) = 0.08/M•s

  14. A product rate = [A] M/s D[A] - M = k [A] Dt [A] = [A]0e−kt ln[A] = ln[A]0 - kt D[A] rate = - Dt First-Order Reactions rate = k [A] = 1/s or s-1 k = [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0

  15. Graphical Determination of k 2N2O5 4NO2 (g) + O2 (g)

  16. The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ? 0.88 M ln 0.14 M = 2.8 x 10-2 s-1 ln ln[A]0 – ln[A] = k k [A]0 [A] [A]0 = 0.88 M ln[A] = ln[A]0 - kt [A] = 0.14 M kt = ln[A]0 – ln[A] = 66 s t =

  17. [A]0 ln t½ [A]0/2 0.693 = = = = k k t½ ln 2 ln 2 0.693 = k k 5.7 x 10-4 s-1 First-Order Reactions The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t½ = t when [A] = [A]0/2 What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s-1)

  18. A product # of half-lives [A] = [A]0/n First-order reaction 1 2 2 4 3 8 4 16

  19. A product rate = [A]2 M/s D[A] 1 1 - M2 = k [A]2 = + kt Dt [A] [A]0 t½ = D[A] rate = - Dt 1 k[A]0 Second-Order Reactions rate = k [A]2 = 1/M•s k = [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2

  20. A product rate [A]0 D[A] - = k Dt [A]0 t½ = D[A] 2k rate = - Dt Zero-Order Reactions rate = k [A]0 = k = M/s k = [A] is the concentration of A at any time t [A] = [A]0 - kt [A]0 is the concentration of A at time t = 0 t½ = t when [A] = [A]0/2

  21. Concentration-Time Equation Order Rate Law Half-Life 1 1 = + kt [A] [A]0 = [A]0 t½ = t½ t½ = ln 2 2k k 1 k[A]0 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions [A] = [A]0 - kt rate = k 0 ln[A] = ln[A]0 - kt 1 rate = k [A] 2 rate = k [A]2

  22. + A + B AB C + D + Exothermic Reaction Endothermic Reaction The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction.

  23. ln k = - + lnA Ea 1 T R Temperature Dependence of the Rate Constant (Arrhenius equation) Eais the activation energy (J/mol) R is the gas constant (8.314 J/K•mol) T is the absolute temperature A is the frequency factor Alternate format:

  24. Alternate Form of the Arrhenius Equation At two temperatures, T1 and T2 or

  25. 2NO (g) + O2 (g) 2NO2 (g) Elementary step: NO + NO N2O2 + Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. N2O2 is detected during the reaction!

  26. 2NO (g) + O2 (g) 2NO2 (g) Mechanism:

  27. Elementary step: NO + NO N2O2 + Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. • The molecularity of a reaction is the number of molecules reacting in an elementary step. • Unimolecular reaction – elementary step with 1 molecule • Bimolecular reaction – elementary step with 2 molecules • Termolecular reaction – elementary step with 3 molecules

  28. Unimolecular reaction Bimolecular reaction Bimolecular reaction A + B products A + A products A products Rate Laws and Elementary Steps rate = k [A] rate = k [A][B] rate = k [A]2 • Writing plausible reaction mechanisms: • The sum of the elementary steps must give the overall balanced equation for the reaction. • The rate-determining step should predict the same rate law that is determined experimentally. The rate-determining step is the sloweststep in the sequence of steps leading to product formation.

  29. Sequence of Steps in Studying a Reaction Mechanism

  30. Step 1: Step 2: NO2 + NO2 NO + NO3 NO2+ CO NO + CO2 NO3 + CO NO2 + CO2 The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps: What is the equation for the overall reaction? What is the intermediate? NO3 What can you say about the relative rates of steps 1 and 2? rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2

  31. Ea k ′ Ea< Ea A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. Uncatalyzed Catalyzed ratecatalyzed > rateuncatalyzed

  32. In heterogeneous catalysis, the reactants and the catalysts are in different phases. • Haber synthesis of ammonia • Catalytic converters In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. • Acid catalysis • Base catalysis

  33. Fe/Al2O3/K2O N2 (g) + 3H2 (g) 2NH3 (g) catalyst Haber Process

  34. catalytic CO + Unburned Hydrocarbons + O2 CO2 + H2O converter catalytic 2NO + 2NO2 2N2 + 3O2 converter Catalytic Converters

  35. Enzyme Catalysis

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