handout19 - 5.6 Reaction Mechanisms BACKGROUND: Imagine we...

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5.6 Reaction Mechanisms BACKGROUND: Imagine we are trying to analyse a complicated chemical process (such as the combustion of wood) E C D I J K M L X Y B Z A We have many different reactions involving different reactants intermediate species products with different orders and rate constants. Typicall questions might be: • What are the reaction rates of the products? • How do the experimental conditions affect product yields? (E.g. combustion of wood under low oxygen conditions charcoal) • What are the rate-limiting steps? 5–25
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Would like to know how reactant and product concentrations change as a function of time (i.e. [ A ]( t ) for species A ) This requires solving differential equations. QUESTION: How can we begin to analyse the overall reaction rate of such a process? ANSWER: Adopt a Bottom-up Approach (need ’building blocks’ + rules for how to combine the building blocks into higher structures). building blocks: need a catalogue of elementary reactions. Elementary reaction : a reaction in which reactants form products via a single transition state. each elementary reaction can be described by an Ordinary Differential Equation (ODE). reaction mechanism: determined by experiment. Reaction Mechanism : list of elementary reactions that occur in the overall process. leads to a set of coupled ODE’s Solve the coupled ODE’s analytically. . . in general, this will not be possible. apply the steady-state approximation to obtain an approximate solution. 5–26
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Numerical Integration of ODE’s PROBLEM: For more complicated reactions (many components, higher orders) the ODE’s rapidly become complex, making an analytic solution difficult. SOLUTION: Use computers to numerically integrate the coupled differential equations. EXAMPLE: Euler’s Method of Integration Let’s assume we would like to solve the following first-order reaction: A k B . (5.54) The differential rate expression reads d [ A ]( t ) d t = - k [ A ]( t ) . (5.55) IDEA: We replace the time derivative by a small time difference to obtain an expression for the change in [ A ]( t ) over a small time interval Δ t : Δ [ A ]( t ) Δ t = - k [ A ]( t ) (5.56) Δ [ A ]( t ) = - k [ A ]( t ) Δ t (5.57) 5–27
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For example, if we know the initial concentration [ A ] 0 , then we can approximate the concentration at a slightly later time t = Δ t : [ A ]( Δ t ) = [ A ] 0 + Δ [ A ]( 0 ) (5.58) = [ A ] 0 + - k [ A ] 0 Δ t (5.59) = [ A ] 0 + h
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This note was uploaded on 12/05/2010 for the course CHBE 251 taught by Professor Scotty during the Winter '09 term at The University of British Columbia.

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handout19 - 5.6 Reaction Mechanisms BACKGROUND: Imagine we...

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