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Topic 16 Kinetics

Topic 16 Kinetics. Rate expressions Reaction mechanism Activation energy. A + B → C + D. 16.1 Rate expression. Change in concentration usually affects the rate of reaction The change in rate isn’t the same for all reactants (A and B)

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Topic 16 Kinetics

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  1. Topic 16 Kinetics • Rate expressions • Reaction mechanism • Activation energy A + B → C + D

  2. 16.1 Rate expression • Change in concentration usually affects the rate of reaction • The change in rate isn’t the same for all reactants (A and B) • Must be determined by experiment. (Change the concentrations of one reactant and hold the others constant) A + B → C + D rate =k*[A]a*[B]b

  3. If a reaction involves the reactants A, B etc => The Rate expression Rate of reaction = - d[A]/dt = k[A]a[B]b k = rate constant • Order of reaction:“a” in substance A and “b” in substance B • Overall order of reaction:a+b

  4. What happens to the rate in the reactionA + B  C + …

  5. Reactants A, B and C in four experiments with altering concentrations:A + B + C => ……

  6. Compare Experiment 1 and 2: Initial rate: [A]: 2x [B] and [C] : constant • 2X rate => [A]1 The reaction is first order in [A]

  7. Compare Experiment 1 and 3: Initial rate: [A]: constant [B]: ½ [C] : constant • same rate => [B]0 The reaction is zero order in [B]

  8. Compare Experiment 2 and 4: Initial rate: [A]: constant [B]: constant [C] : 3X • 33 = 9X=> [C]2 The reaction is second order in [C]

  9. Conclusion • Rate = k*[A]1’*[B]0*[C]2= k*[A]1*[C]2 • Overall order 1+2 = 3 • k can be calculated using the data from one of the experiments above

  10. Exercises • 1-2 on page 120

  11. The order can also be found in a graph where initial concentration is set against initial rate.The gradient of the graph => rate of the reaction.

  12. First order reactionsThey show an exponential decrease:the time to half the concentration is equal to go from ½ to 1/4Half life, t½ = 0.693/k

  13. 16.2 Reaction mechanism Types of reactions: Molecularity • A  Products Unimolecular • A + B  Products Bimolecular • In a Bimolecular reaction the reactants collide and form an activated complex

  14. 2 molecules Nucleophilic Substitution bimolecular, SN2- topic 10

  15. If we study the reaction: CH3COCH3+ I2CH3COCH2I + HI H+ • It could be a bimolecular process with a rate expression rate = k*[CH3COCH3] *[I2] • The rate is independent of [I2], but first-order in propanone and acid => rate = k*[CH3COCH3]*[H+] • The reaction must proceed through a series of steps, a mechanism must be found:

  16. Intermediate CH3C(OH+)CH3 is known as a intermediate, not an activated complex, though it occur at an energy minimum. In the mechanism there will be several activated complexes

  17. Activated complex = Transition state, T

  18. Exercises • 1 and 2 on page 122

  19. 16.3 Activation energy Recall: Maxwell-Boltzmann energy distribution curve. Temperature  Average speed Higher temperature =>More particles with higher speed => Greater proportion of particles with energy enough to react

  20. The Arrhenius equation • The rate constant, k, can be given if collision rate and orientation is given • Ea = activation energy • T = temperature, K • R = Gas constant

  21. The equation can also be given in a logarithmic form:

  22. Exersize: Consider the following graph of lnk against (temperature in Kelvin) for the second order decomposition of N2O into N2 and O N2O → N2 + O (a) State how the rate constant, k varies with temperature, T (b) Determine the activation energy, Ea, for this reaction. (c) The rate expression for this reaction is rate = k [N2O]2 and the rate constant is0.244 dm3mol–1 s–1 at 750 °C. A sample of N2O of concentration 0.200 moldm–3 is allowed to decompose. Calculate the rate when 10 % of the N2O has reacted.

  23. Solution: (a) State how the rate constant, k varies with temperature, T The Arrhenius equation (it’s in the Data booklet): k=Ae(-Ea/RT) Logaritming the equation on both sides: lnk=lnA –Ea/RT In the graphweseethat the gradient is negative Answer: when k increases T decreases (b) Determine the activation energy, Ea, for this reaction The gradient (-Ea/R) can be calculated from the graph: DY/DX= -3/0,1*10-3= -3*104= -30000 Therefore: Ea=gradient*R=-30000*8.31= 2,49*105 Answer: Ea= 2,49*105

  24. Solution: (c) The rate expression for this reaction is rate = k [N2O]2 and the rate constant is0.244 dm3mol–1 s–1 at 750 °C. A sample of N2O of concentration 0.200 moldm–3 is allowed to decompose. Calculate the rate when 10 % of the N2O has reacted. 10 % has reacted → 90 % left → 0.200*0.9= 0.180 mol/dm3 Rate= 0.244 * [0.180]2= 7.91*10-3

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