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Chemical Kinetics

Chemical Kinetics. Chapter 14. a A + b B c C + d D. The Rate Law. Rate law – description of the effect of concentration on rate. Rate = k [A] x [B] y. reaction is x th order in A. reaction is y th order in B. reaction is (x +y)th order overall.

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Chemical Kinetics

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  1. Chemical Kinetics Chapter 14

  2. aA + bB cC + dD The Rate Law Rate law – description of the effect of concentration on rate Rate = k [A]x[B]y reaction is xth order in A reaction is yth order in B reaction is (x +y)th order overall

  3. F2(g) + 2ClO2(g) 2FClO2(g) 1 Rate Laws • Rate laws always determined experimentally • Reaction order always defined in terms of reactant (not product) concentrations • Order of a reactant is notrelated to the stoichiometric coefficients rate = k [F2][ClO2]

  4. N2(g) + 2 H2O(l) NH4+(aq) + NO2−(aq) Concentration and Initial Rates Table 14.2 Assume: rate = k [NH4+]x[NO2−]y Double [NH4+] with [NO2−] constant: Rate doubles x = 1 Double [NO2−] with [NH4+] constant: Rate doubles y = 1 Therefore rate law is rate = k [NH4+][NO2−]

  5. 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

  6. Relation Between Concentration and Time • Rate law provides rate as a function of concentration • Need relationship between concentration and time : • i.e., How do we determine the concentration of a • reactant at some specific time?

  7. A product rate = [A] M/s D[A] − M = k [A] Dt D[A] rate = − Dt First-Order Reactions rate = k [A] = 1/s or s-1 k = Which now integrates to: [A] is the concentration of A at any time t ln[A] = ln[A]o - kt [A]0 is the concentration of A at time t=0 Integrated rate law

  8. First-Order Reactions Consider the process in which methyl isonitrile is converted to acetonitrile Fig 14.7(a) Data collected for this reaction at 198.9 °C. CH3NC CH3CN

  9. First-Order Reactions Fig 14.7 • Plot of ln P vs time, results in a straight line • Therefore, • Process is first-order • k is the negative of the slope: 5.1  10-5 s−1

  10. Decomposition of N2O5 2 N2O5(in CCl4)→4 NO2(g) + O2(g) orange red-brown →

  11. Decomposition of N2O5 2 N2O5(in CCl4)→ 4 NO2(g) + O2(g) Plot of ln [N2O5] vs time Linear plot indicates: 1st order

  12. Second-Order Reactions A + B product A product D[A] 1 1 − = k [A]2 = + kt Dt [A] [A]o D[A] rate = − Dt One type: Initial rate law: rate = k [A]2 Combining the two rate expressions: Integrated rate law Which now integrates to: [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 Second type: rate = k [A][B]

  13. NO2 (g) NO (g) + 1/2 O2 (g) 1 1 = + kt [A] [A]o Second-Order Reactions The decomposition of NO2 at 300°C is described by the equation Fig 14.8 plot of 1/[NO2] vs time: plot of ln[NO2] vs time:

  14. t½ = 1 k[A]o Half-Life Fig 14.9 • Half-life = time required for one-half of a reactant to react • t½ = t when [A] = [A]0/2 • Because [A] at t1/2 is one-half of the original [A], [A]t = 0.5 [A]0 • 1st order t1/2 = 0.693/k • 2nd order 1st order rxn

  15. Concentration-Time Equation Order Rate Law Half-Life 1 1 = + kt [A] [A]o = [A]o t½ = t½ t½ = ln2 2k k 1 k[A]o Summary of the Kinetics of Reactions [A] = [A]o - kt rate = k 0 ln[A] = ln[A]o - kt 1 rate = k [A] 2 rate = k [A]2

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