handout20 - 5.8 Solving Reaction Mechanisms REMINDER Our...

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5.8 Solving Reaction Mechanisms REMINDER: Our task, as outlined on page 5–26, is to solve a set of coupled ODE’s. AIM: Introduce some tricks of the trade that can be used to solve coupled ODE’s analytically. Also introduce some interesting cases as we go along. Steady-state 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 ODE’s. 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 B k 2 C (5.132) where both reactions are first order. We would have to solve the coupled ODE’s: - d [ A ] d t = k 1 [ A ]( t ) (5.133) 5–46
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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 ] 0 e - k 1 t . (5.136) Solving for [ B ]( t ) is not so easy; Apply the steady state approximation (SSA) d [ B ] d t = 0 So then we write 0 = 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 ] 0 e - k 1 t . (5.142) The solution to this is [ C ] SS ( t ) = [ A ] 0 1 - e - k 1 t . (5.143) 5–47
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Now we can compare these approximate results to the accurate ones (Eqs 5.129 =5.130): k 2 k 1 0 20 40 60 80 100 120 concentration time [A] [B] [C] [A] SS [B] SS [C] SS k 2 >> k 1 0 20 40 60 80 100 120 concentration time [A] [B] [C] [A] SS [B] SS [C] SS 5–48
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k 2 << k 1 0 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 Steady-state 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 ODE’s for a reversible reaction A k f 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) 5–49
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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. So from Thermodynamics, we know what the equilibrium concentrations [ A ] ( ) and [ B ] ( ) will be; Now we ask: how fast is this state of dynamic equilibrium reached?
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