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Chemistry 232

Chemistry 232. Chemical Kinetics. Chemical Kinetics. Chemical kinetics - speed or rate at which a reaction occurs How are rates of reactions affected by Reactant concentration? Temperature? Reactant states? Catalysts?. The Instantaneous Reaction Rate. Consider the following reaction

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Chemistry 232

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  1. Chemistry 232 Chemical Kinetics

  2. Chemical Kinetics • Chemical kinetics - speed or rate at which a reaction occurs • How are rates of reactions affected by • Reactant concentration? • Temperature? • Reactant states? • Catalysts?

  3. The Instantaneous Reaction Rate • Consider the following reaction A + B  C • Define the instantaneous rate of consumption of reactant A, A

  4. Reaction Rates and Reaction Stoichiometry • Look at the reaction O3(g) + NO(g) ® NO2(g) + O2(g)

  5. Another Example 2 NOCl (g)  2 NO + 1 Cl2 (g) WHY? 2 moles of NOCl disappear for every 1 mole Cl2 formed.

  6. The General Case a A + b B ®c C + d D rate = -1 d[A] = -1 d[B] = +1 d[C] = +1 d[D] a dt b dt c dt d dt • Why do we define our rate in this way? • removes ambiguity in the measurement of reaction rates • Obtain a single rate for the entire equation, not just for the change in a single reactant orproduct.

  7. Alternative Definition of the Rate • Rate of conversion related to the advancement of the reaction, . V = solution volume

  8. An Example • Examine the following reaction 2 N2O5 (g)  4 NO2 (g) + O2 (g)

  9. The N2O5 Decomposition Note – constant volume system

  10. The Rate Law • Relates rate of the reaction to the reactant concentrations and rate constant • For a general reaction a A + b B + c C ® d D + e E

  11. Rate Laws (Cont’d) • The only way that we can determine the superscripts (x, y, and z) for a non-elementary chemical reaction is by experimentation. • Use the isolation method (see first year textbooks).

  12. For a general reaction  x + y + z = reaction order e.g. X = 1; Y = 1; Z = 0 2nd order reaction (x + y + z = 2) X = 0; Y = 0; Z = 1 (1st order reaction) X = 2; Y = 0; Z = 0 (2nd order)

  13. Rate Laws for Multistep Processes • Chemical reactions generally proceed via a large number of elementary steps - the reaction mechanism • The slowest elementary step Þ the rate determining step (rds).

  14. An Example Reaction Mechanism • O atoms are intermediates in the above reaction sequence. • Intermediates – generally small, indeterminate concentrations.

  15. Integrated Rate Laws • The rate law gives us information about how the concentration of the reactant varies with time • How much reactant remains after specified period of time? Use the integrated rate laws.

  16. First Order Reaction A  product Rate = v = - d[A]/dt = k[A] • How does the concentration of the reactant depend on time? k has units of s-1

  17. The Half-life of a First Order Reaction • For a first order reaction, the half-life t1/2 is calculated as follows.

  18. Radioactive Decay • Radioactive Samples decay according to first order kinetics. • This is the half-life of samples containing e.g. 14C , 239Pu, 99Tc. • Example

  19. Second Order Reaction A ® products v = k[A]2 A + B ® products v = k[A][B] • Case 1 is 2nd order in A • Case 2 is 1st order in A and B and 2nd order overall

  20. The Dependence of Concentration on Time • For a second order process where v = k[A]2

  21. Half-life for This Second Order Reaction. • [A] at t = t½ = ½ [A]0

  22. Other Second Order Reactions • Examine the Case 2 from above A + B ® products v = k[A][B]

  23. A Pseudo-first Order Reaction • Example hydration of methyl iodide CH3I(aq) + H2O(l)  CH3OH(aq) + H+(aq) + I-(aq) Rate = k [CH3I] [H2O] • Since we carry out the reaction in aqueous solution [H2O] >>>> [CH3I] \ [H2O] doesn’t change by a lot

  24. Pseudo-first Order (cont’d) • Since the concentration of H2O is essentially constant v = k [CH3I] [constant] = k`[CH3I] where k` = k [H2O] • The reaction is pseudo first order since it appears to be first order, but it is actually a second order process.

  25. Sequential First Order Reactions • Suppose we have two first order reactions occurring in sequence.

  26. What is Our Intermediate? • B is an intermediate in the above reaction sequence. • Clearly B is formed in the first elementary step of the reaction mechanism and consumed in the final step.

  27. The Concentration Dependencies of the Species • The amounts of the reactants are related to the reaction rates as follows.

  28. Sequential Reactions (3) • For a set of initial conditions [A]o 0, and [B]o = [C]o =0 mol/L.

  29. Sequential Reactions (4) • The concentration of the intermediate can be written

  30. Sequential Reactions (5)

  31. The Rate Determining Step of the Reaction • What happens if one of the steps in the reaction is much slower than the other reaction step? Note – assuming ka <<< kb

  32. The Rate Determining Step of the Reaction (2) • If kb <<< ka

  33. Temperature Dependence of Reaction Rates • Reaction rates generally increase with increasing temperature. • Arrhenius Equation A = pre-exponential factor Ea = the activation energy

  34. Reactions Approaching Equilibrium • Examine the concentration profiles for the following simple process. A ⇌ B

  35. Approaching Equilibrium • Calculate the amounts of A and B at equilibrium.

  36. The Equilibrium Condition • At equilibrium, vA,eq = vB,eq.

  37. Elementary steps and the Molecularity • Kinetics of the elementary step depends only on the number of reactant molecules in that step! • Molecularity  the number of reactant molecules that participate in elementary steps

  38. The Kinetics of Elementary Steps • For the elementary step • unimolecular step • For elementary steps involving more than one reactant • a bimolecular step

  39. The Kinetics of Elementary Steps • For the elementary step • unimolecular step • For elementary steps involving more than one reactant • a bimolecular step

  40. For the step • a termolecular (three molecule) step. • Termolecular (and higher) steps are not that common in reaction mechanisms.

  41. The Steady-State Approximation • Examine the following simple reaction mechanism Rate of product formation, vp, is proportional to the concentration of an intermediate.

  42. Applying the Steady State Approximation (SSA) • Look for the intermediate in the mechanism. • Step 1 – B is produced. • Reverse of Step 1 – B is consumed. • Step 2 – B is consumed.

  43. The SSA (Cont’d) • The SSA applied to the intermediate B.

  44. SSA – The Final Step • Substitute the expression for the concentration of B into the rate law vp.

  45. Competing Reactions • Imagine a reaction with a competing side reaction.

  46. The Reaction “Yield” • We can calculate the amount of material produced from the competing reactions. • kJ = the rate constant for the reaction J.

  47. Activated Complex Theory • Consider the following bimolecular reaction • Presume that the reaction proceeds through the transition state?

  48. The Activated Complex • The species temporarily formed by the reactant molecules – the activated complex. • A small fraction of molecules usually have the required kinetic energy to get to the transition state • The concentration of the activated complex is extremely small.

  49. Transition State Theory (TST) • TST pictures the bi-molecular reaction proceeding through the activated complex in a rapid-pre-equilibrium.

  50. TST (II) • From the thermodynamic equilibrium constant for the formation of AB↕

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