Week 5.1 - Compact Notation Derivation for a simple Mechanism
Compact Notation Derivation for a Simple Mechanism
Objective:
To derive the reaction rate ODEs of a simple reaction mechanism learnt in the chemical kinetics lectures.
Reaction Mechanism i
CO + O2 ⇌ CO2 + O R1
O + H2o ⇌ OH + OH R2
CO + OH ⇌ CO2 + H R3
H + O2 ⇌ OH + O R4
j=species
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| CO | O2 | CO2 | O | H2O | OH | H |
kf=forward reaction coefficient kr=reverse reaction coefficient
`d[[CO] ]/dt =-kf1[CO][o2]+kr1[CO2][O]-kf3[CO][OH]+kr3[CO2][H]`
`d[[O2]]/dt=-kf1[CO][O2]+kr1[CO2][O]-kf4[h][O2]+kr3[OH][O]`
`d[[CO2]]/dt=kf1[CO][O2]-kr1[CO2][O]+kf3[CO][OH]-kr3[CO2][H]`
`d[[O]]/dt=kf1[CO][O2]-kr1[CO2][O]-kf2[O][H2O]+kr2[OH][OH]+kf4[H][O2]-kr4[OH][O]`
`d[[H2O]]/dt=-kf2[O][H2O]+kr2[OH][OH]`
`d[[OH]]/dt=2*(kf2[O][H2O]-kr2[OH][OH])-kf3[CO][OH]+kr3[CO2][H]+kf4[H][O2]-kr4[OH][O]`
`d[[H]]/dt=kf3[CO][Oh]-kr3[CO2][H]-kf4[H][O2]+kr4[OH][O]`
Defining the reactant matrix
vji'=`[[1,1,0,0,0,0,0],[0,0,0,1,1,0,0],[1,0,0,0,0,1,0],[0,1,0,0,0,0,1]]`
product matrix
vji''=`[[0,0,1,1,0,0,0],[0,0,0,0,0,2,0],[0,0,1,0,0,0,1],[0,0,0,1,0,1,0]]`
net reaction rate
vji=vji''-vji=`[[-1,-1,1,1,0,0,0],[0,0,0,-1,-1,2,0],[-1,0,1,0,0,-1,1],[0,-1,0,1,0,1,-1]]`
we have
`i=(kfi`j=1 prod^N [Xj]^(Vji' ))-(kri j=1 prod^N[Xj]^(Vji'')`
there fore
from i 1 to 4
`q1=kfq[CO]^1[O2]^1-kr1[CO2]^1[O]^1`
`q2=kf2[O]^1[H2O]^1-kr2[OH]^2`
`q3=kf3[CO]^1[OH]^1-kr3[CO2]^1[H]^1`
`q4=kf4[O2]^1[H]^1-kr4[O]^1[OH]^1`
since for each species there is an production rate
then the rate of concentration change of a particular species is given by
`Wj=d[[Xj]]/dt`
by summing up
`Wj=sum^L(i=1)vj^tqi`
where L is the number of reaction
we will get a stiff ODE system
`((w1),(w2),(w3),(w4),(w5),(w6),(w7))=[[-1,0,-1,0],[-1,0,0,-1],[1,0,1,0],[1,-1,0,1],[0,-1,0,0],[0,2,-1,-1],[0,0,1,-1]]*((q1),(q2),(q3),(q4))`
w1=`d[[CO]]/dt=-q1-q3`
`d[[CO] ]/dt =-kf1[CO]^1[O2]^1+kr1[CO2]^1[O]^1-kf3[CO]^1[OH]^1+kr3[CO2]^1[H]^1`
w2=`d[[O2]]/dt=-q1-q4`
`d[[O2]]/dt=-kf1[CO]^1[O2]^1+kr1[CO2]^1[O]^1-kf4[H]^1[O2]^1+kr3[OH]^1[O]^1`
w3=`d[[Co2]]/dt=q1+q3`
`d[[CO2]]/dt=kf1[CO]^1[O2]^1-kr1[CO2]^1[O]^1+kf3[CO]^1[OH]^1-kr3[CO2]^1[H]^1`
w4=`d[[O]]/dt=q1-q2+q4`
`d[[O]]/dt=kf1[CO]^1[O2]^1-kr1[CO2]^1[O]^1-kf2[O]^1[H2O]^1+kr2[OH]^2+kf4[H]^1[O2]^1-kr4[OH]^1[O]^1`
w5=`d[[H2O]]/dt=-q2`
`d[[H2O]]/dt=-kf2[O]^1[H2O]^1+kr2[OH]^2`
w6=`d[[OH]]/dt=2q2-q3+q4`
`d[[OH]]/dt=2*(kf2[O]^1[H2O]^1-kr2[OH]^2)-kf3[CO]^1[OH]^1+kr3[CO2]^1[H]^1+kf4[H]^1[O2]^1-kr4[OH]^1[O]^1`
w7=`d[[H]]/dt=q3-q4`
`d[[H]]/dt=kf3[CO]^1[OH]^1-kr3[CO2]^1[H]^1-kf4[H]^1[O2]^1+kr4[OH]^1[O]^1`
Conclusion;
Since this ODE is a stiff ODE we need implicit scheme to solve while explicit scheme like euler won't work out
By following this method we can predict reaction pathways by deriving the ODE Reaction rate