Challenge

Week 5 - Compact Notation Derivation for a simple Mechanism

Reaction Mechanism

CO + O2 -> CO2 + O ......1

O + H2o -> OH + OH .....2

CO + OH -> CO2 + H . .....3

H + O2 -> OH + O .....4

i = Reaction number i.e ( 1, 2,3,4 )

j Species
1 O
2 O2
3 CO
4 CO2
5 H
6 OH
7 H2O

we can write matrix for reactant and product side

for reactant side

`V_(ji)^' = [[0,1,1,0,0,0,0],[1,0,0,0,0,0,1],[0,0,1,0,0,1,0],[0,1,0,0,1,0,0]]`

for product side

`V_(ji)^('') = [[1,0,0,1,0,0,0],[0,0,0,0,0,2,0],[0,0,0,1,1,0,0],[1,0,0,0,0,1,0]]`

so we can write

`v_(ji) = v_(ji)^('') - v_(ji)^'`

`v_(ji) = [[1,-1,-1,1,0,0,0],[-1,0,0,0,0,2,-1],[0,0,-1,1,1,-1,0],[1,-1,0,0,-1,1,0]]`

we know

`q_i = k_(fi) xx prod_(i=1)^7[X_i] - k_(ri) xx prod_(i=1)^7[X_i]`

we can write for our above four reaction,

`q_1 = k_(f1) [CO][O2]-k_(r1)[CO2][O]`

`q_2 = k_(f2)[O][H2O] - k_(r2)[OH][OH]`

`q_3 = k_(f3)[CO][OH]-k_(r3)[CO2][H]`

`q_4 = k_(f4) [H][O2]-k_(r3)[OH][O]`

we can write `k_(ri) = 0` if the reaction is only forwared reaction ( i.e in only in one direction )

`omega_j = sum_(i=0)^L v_(ji) q_i`

whare L is the number of reactions

`omega` is net production rate of particular species

for species O i.e ( j = 1)

`omega_1 = v_11 q_1 + v_12q_2+v_13q_3+v_14q_4`

from above matrix we can write

`v_11 = 1 , v_12 = -1 , v_13 = 0 , v_14 = 1`

`:.omega_1 = q_1 -q_2 +q_3`

similarly

`omega_2 = -q_1 -q_4`

`omega_3 = -q_1 -q_3`

..

..

..

..

`omega_7 = -q_2`

we can write in matrix foam as

`[[omega_1],[omega_2],[omega_3],[omega_4],[omega_5],[omega_6],[omega_7]] = [[1,-1,0,1],[-1,0,0,-1],[-1,0,-1,0],[1,0,1,0],[0,0,1,-1],[0,2,-1,1],[0,-1,0,0]]xx[[q_1],[q_2],[q_3],[q_4]]`

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