Week 5.1 - Compact Notation Derivation for a simple Mechanism

Aim: The objective of this challenge is need to derive the reaction rate ODEs & production of each species for the given reaction mechanism.

Solution: First, need to start from finding how much of species is available in the given reaction mechanism. Here, there is 4 reactions are given. So,

1. Reactions (i): The first given reaction mechanism is called as R1 and in order so call the reactions as R2, R3 & R4.
2. Species (j): The available species in those given reaction are 7 species. They are CO, O₂, CO₂, O, H₂O, OH & H.

Now, we have to obtain the stoichiometric co-efficient of the species from the given reactions for both the reactant and product in the matrix format (species in row & reactions in column).

Reactant Stoichiometric co-efficient(V_ji^’):

V_ji^’ = `[[j/i,CO,O_2,CO_2,O,H_2O,OH,H],[R_1,1,1,0,0,0,0,0],[R_2,0,0,0,1,1,0,0],[R_3,1,0,0,0,0,1,0],[R_4,0,1,0,0,0,0,1]]`

Product Stoichiometric co-efficient (V_ji^’’):

V_ji^’’ = `[[j/i,CO,O_2,CO_2,O,H_2O,OH,H],[R_1,0,0,1,1,0,0,0],[R_2,0,0,0,0,0,2,0],[R_3,0,0,1,0,0,0,1],[R_4,0,0,0,1,0,1,0]]`

Total Stoichiometric co-efficient (V_ji):

V_ji = V_ji^’’ - V_ji^’ =`[[j/i,CO,O_2,CO_2,O,H_2O,OH,H],[R_1,-1,-1,1,1,0,0,0],[R_2,0,0,0,-1,-1,2,0],[R_3,-1,0,1,0,0,-1,1],[R_4,0,-1,0,1,0,1,-1]]`

Now, we need to calculate the net reaction rate

`q_i` = k_fi * `π_1^7 ** [X]_i` – k_ri * `π_1^7 ** [X]_i`

Where

k_fi is the rate of forward reaction

k_ri is the rate of reverse reaction

q_i is the net reaction

Apply this equation for each given reaction

q₁ = k_f1 * [CO]*[O₂] – k_r1 * [CO₂]*[O]

q₂ = k_f2 * [O]*[H₂O] – k_r2 * [OH]*[OH]

q₃ = k_f3 * [CO]*[OH] – k_r3 * [CO₂]*[H]

q₄ = k_f4 * [H]*[O₂] – k_r4 * [OH]*[O]

Now, calculate the product rate

`ω_j = sum_(i=1)^L`V_{ji *} q_i

where, L is the length of the reaction

Apply this equation for each given reaction

`ω_1 = (V_11 ** q_1) + (V_21 ** q_2) + (V_31 ** q_3) + (V_41 ** q_4)`

`ω_2 = (V_21 ** q_1) + (V_22 ** q_2) + (V_23 ** q_3) + (V_24 ** q_4)`

`ω_3 = (V_31 ** q_1) + (V_32 ** q_2) + (V_33 ** q_3) + (V_34 ** q_4)`

.

.

.

`ω_7 = (V_71 ** q_1) + (V_72 ** q_2) + (V_73 ** q_3) + (V_74 ** q_4)`

Now, form the matrix from the above equations

`[(ω_1),(ω_2),(ω_3),(ω_4),(ω_5),(ω_6),(ω_7)] = [(-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))`

The final solution of the product of each species for given reactions are

`ω_1 = -q_1 - q_3`

`ω_2 = -q_1 - q_4`

`ω_3 = q_1 + q_3`

`ω_4 = q_1 - q_2 + q_4`

`ω_5 = -q_2`

`ω_6 = 2q_2 - q_3 – q_4`

`ω_7 = q_3 – q_4`