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Chemical kinetics

Chemical kinetics. The rate of a chemical reaction is dependent on:. reactant concentrations state of reactants (solid, liquid, powder, etc.) temperature (e.g., eggs cook faster at higher temperatures)

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Chemical kinetics

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  1. Chemical kinetics The rate of a chemical reaction is dependent on: • reactant concentrations • state of reactants (solid, liquid, powder, etc.) • temperature (e.g., eggs cook faster at higher temperatures) • catalyst (e.g., “catalytic converter” in your car speeds the formation of less polluting products from your engine) Kinetics predicts on how fast you reach equilibrium... not the extent to which the reaction proceeds! 

  2. instantaneous reaction rate (slope at a given instant in time) Reaction rates are determined experimentally NOTE: Rate varies with time 

  3. Average Reaction Rate

  4. Relative Rates • aA + bB → cC + dD • Rate =

  5. Example • The reaction of hydrogen with nitrogen to produce ammonia is 3H2 + N2 2NH3 • If the rate of appearance of NH3 is 3.0 x 10-6 M/s, what is the rate of disappearance of H2?

  6. Write relative rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of the products 3O2(g) → 2O3(g) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

  7. Types of Rate Laws Differential Rate Law: expresses how rate depends on concentration. Integrated Rate Law: expresses how concentration depends on time.

  8. units of M/s (moles/L-s) Rate constant (units vary) • First order in B • Second order in A • Overall order of reaction = 1 + 2 = 3 • Changing the concentration of A will have much bigger impact on reaction rate than will changing the concentration of B A and B are reactants Differential Rate Law This rate law shows how the rate of a reaction depends on concentrations of various species in the reaction. The Rate Law is determined experimentally & does not necessarily reflect the stoichiometry of the reaction 

  9. How does value of k affect reaction rate?

  10. Rate Law Example • The reaction2NO(g) + 2H2(g)  N2(g) + 2H2O(g)has the rate law R = k [NO]2 [H2] • Therefore the reaction is 2nd order in NO, 1st order in H2 and 3rd order overall.

  11. Rate Constant Units • The units for the rate constant depend on the rate law. • For example, if the rate law is R = k [NO]2 [Br2]then the units are: • R  M sec-1 • therefore, k

  12. Measured Reaction Rates • The rate of reaction changes with time, as the concentrations of reactants and products change. • The rate constant, but not the rate, is independent of time. • The initial rate is the reaction rate measured before the initial concentrations have had time to change.

  13. Determining Reaction Orders • The rate law for a given reaction must be determined experimentally. This means determining the order with respect to each species involved. • The two major methods are • Method of Initial Rates • Integrated Rate Law Method

  14. Method of Initial Rates

  15. Method of Initial Rates ….involves measuring the rate of reaction at very short times before any significant changes in concentration occur

  16. The initial rate is determined in several experiments using different initial concentrations. e.g. BrO3- +5Br- + 6H3O+ 3Br2 +_ 9H2O initial concentrations in starting solution 

  17. may be able to use this formula useful for non-integer orders How to use the initial rate data Note: k does not change 

  18. return to problem... BrO3- Find exp’t pair where only ONE conc changes Find similar pair of exp’ts for other reactants Br- H3O+ Rate=k[BrO3-][Br-][H3O+]2 

  19. Integrated Rate Laws

  20. First-Order Rate Law • For aA  products in a second-order reaction,

  21. Half lives, t1/2 When [A]t=1/2 [A]0 then t = t1/2 Can we predict concentration with time? We will consider only first order reaction... Rate = k[A] AB 

  22. Example problem: radioactive decay (a first order process) 14C has a half-life of 5,730 years. If the concentration of 14C originally in an artifact was 1.3x10-9 moles/g and the current concentration is 0.87x10-9 moles/g,. How old is the artifact? 

  23. Recognizing First-Order • Plot the experimental ln [A] vs. time • If the graph is linear, the reaction is first-order. • The rate constant is k = -slope ln [A] = - kt + ln [A]0 y = m t + b

  24. SO2Cl Example • The reaction SO2Cl2(g)  SO2(g) + Cl2(g) gives the following experimental data:

  25. SO2Cl Example ln P vs time slope = 2.19 x 10-5 sec-1 ln(p/atm) time/sec

  26. Second-Order Rate Law • For aA  products in a second-order reaction,

  27. Second-Order Rate Law (Type I) k 2A  P

  28. Recognizing Second-Order • Plot the experimental 1/[A] vs. time • If the graph is linear, the reaction is second-order. • The rate constant is k = slope y = m t + b

  29. NO2 Example The following data were obtained for the gas-phase decomposition of NO2(g) at 300°C.

  30. First Order Plot ln [NO2] time / s

  31. Second Order Plot 1/ [NO2] time / s

  32. Zero Order Rate Law • Reaction rate is does not depend on concentration • Rate = k • Plot of [A] vs. time produces straight line

  33. Zero Order Rate Law

  34. Reaction Mechanisms

  35. Reaction Mechanism • The series of stepsby which a chemical reaction occurs. • A chemical equation does not tell us how reactants become products - it is a summary of the overall process.

  36. Reaction Mechanism • The mechanism of a reaction is the sequence of individual collisions, known as elementary steps, that take the reactant molecule(s) to the product molecule(s).

  37. Often Used Terms • Elementary Step: A reaction whose rate law can be written from its chemical equation.

  38. Often Used Terms • Intermediate: formed in one step and used up in a subsequent step and so is never seen as a product.

  39. Often Used Terms • Molecularity: the number of species that must collide to produce the reaction indicated by that step.

  40. Ozone Example • The conversion of ozone O3 to oxygen O2 has the overall balanced equation: • 2O3(g)  3O2(g). • A possible mechanism for this reaction has two elementary steps: • O3(g)  O(g) + O2(g) • O3(g) + O(g) 2O2(g) • Net result: 2O3(g)  3O2(g) • O(g) is an intermediate

  41. Reaction Mechanisms Criteria • Elementary steps must add to give overall balanced equation • Rate law of slow step must agree with actual rate law

  42. Rate Determining Steps • Some reactions have a mechanism in which one of the elementary steps is much slower than the others. • Such a slow elementary step is called the rate determining step, since it acts as the bottleneck for the overall reaction. • The overall rate law is then simply the rate law for the rate determining step.

  43. NO + NO  N2O2 N2O2 + O22NO2 (slow) Schematic of rate of reaction for2NO + O2 2NO2 

  44. When the slow step isn’t the first step…intermediates in the rate law Step 1: NO + NO  N2O2 Step 2: N2O2 + O22NO2 (slow)

  45. Collision Model • Molecules must collide with sufficient energy to react. • only a small fraction of collisions produces a reaction. • Arrhenius: An activation energy, Ea,must be overcome.

  46. EFFECT OF TEMPERATURE ON REACTION KINETICSMinimum energy needed for “effective collision”, i.e., product formed no reaction Products 

  47. Temperature dependence of the number of molecules that have the “minimum energy” to react. Activation energy At higher temperatures, more molecules have the minimum energy and the rate of reaction increases. 

  48. An activation barrier present for an exothermic reaction would explain the behavior of such reactions as H2 +O2 Larger Ea slows rate of reaction Activation Energy and rate of reaction 

  49. Arrhenius Equation • Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy Ea). • Orientation of reactants must allow formation of new bonds. (frequency factor, A)

  50. The frequency factor (A) can be visualized as a “geometry factor”. Not all collisions (even with enough energy) will result in a successful reaction. What if it hits in the “wrong spot”?

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