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Unformatted text preview: 5.8 Solving Reaction Mechanisms REMINDER: Our task, as outlined on page 526, is to solve a set of coupled ODEs. AIM: Introduce some tricks of the trade that can be used to solve coupled ODEs analytically. Also introduce some interesting cases as we go along. Steadystate Approximation IDEA: Reactive intermediate species I are being created and consumed during the reaction. We make the simplifying assumption that [ I ] does not change much: d [ I ] d t = 0 . (5.131) use this to simplify the coupled ODEs. Often, the differential equations are reduced to algebraic equations which are easier to solve. EXAMPLE: Lets look at a sequential reaction to illustrate this. REMINDER: We had A k 1 GGGGGA B k 2 GGGGGA C (5.132) where both reactions are first order. We would have to solve the coupled ODEs: d [ A ] d t = k 1 [ A ]( t ) (5.133) 546 d [ B ] d t = k 1 [ A ]( t ) k 2 [ B ]( t ) (5.134) d [ C ] d t = k 2 [ B ]( t ) . (5.135) Lets see if we can find an approximate solution to this problem. Solving for [ A ]( t ) is easy because there is only an [ A ]dependency: [ A ]( t ) = [ A ] e k 1 t . (5.136) Solving for [ B ]( t ) is not so easy; Apply the steady state approximation (SSA) d [ B ] d t = So then we write = d [ B ] d t = k 1 [ A ]( t ) k 2 [ B ] SS ( t ) (5.137) [ B ] SS ( t ) = k 1 k 2 [ A ]( t ) . (5.138) fi for [ C ] we can write: d [ C ] d t = k 2 [ B ] SS ( t ) (5.139) = k 2 k 1 k 2 [ A ]( t ) (5.140) = k 1 [ A ]( t ) (5.141) = k 1 [ A ] e k 1 t . (5.142) The solution to this is [ C ] SS ( t ) = [ A ] 1 e k 1 t . (5.143) 547 Now we can compare these approximate results to the accurate ones (Eqs 5.129 =5.130): k 2 k 1 20 40 60 80 100 120 concentration time [A] [B] [C] [A] SS [B] SS [C] SS k 2 >> k 1 20 40 60 80 100 120 concentration time [A] [B] [C] [A] SS [B] SS [C] SS 548 k 2 << k 1 50 100 150 200 250 300 concentration time [A] [B] [C] [A] SS [B] SS [C] SS The approximation is particularly good when k 2 >> k 1 because d [ B ] d t really is small. The Steadystate Approximation (SSA) is valid when an intermediate species is always low in concentration. This is true when its rate of creation is slow compared to its rate of consumption. Treating Reversible Reactions We are interested in solving the ODEs for a reversible reaction A k f GGGGGB FGGGGG k b B . (5.144) REMINDER: (Eq. 5.45) If the system is at dynamic equilibrium, the two rates must be equal k f [ A ] = k b [ B ] (5.145) 549 and we can write k f k b = thermodynamics z } { = K C [ B ] [ A ] eq  {z } kinetics (5.146) which combines the fields of Kinetics and Thermodynamics....
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 Winter '09
 scotty
 Reaction

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