Week 5.1 - Compact Notation Derivation for a simple Mechanism

Objective : To derive the reaction rate of ODEs & production rate of each species for the given simple reaction mechanism

Reaction Mechanism :

CO+ O2 ⇌ CO2 + O

O + H2O ⇌ OH + OH

CO + OH ⇌ CO2 + H

H + O2 ⇌ OH + O

For determining the rate of reaction , we need to compute forward and backward reaction coefficients which are denoted by k_f and k_r.The positive and negative sign depends on whether the species are on reactants side or products side.

d[CO]/dt = -k_f1[CO][O2] + k_r1[CO2][O] - k_f3[CO][OH] + k_r3[CO2][O]

d[O2]/dt = -k_f1[CO][O2] + k_r1[CO2][O] - k_f4[H][O2] + k_r4[OH][O]

d[CO2]/dt = k_f1[CO][O2] - k_r1[CO2][O] + k_f3[CO][OH] - k_r3[CO2][H]

d[O]/dt = k_f1[CO][O2] - k_r1[CO2][O] -k_f2[O][H2O] + k_r2[OH][OH] + k_f4[H][O2] - k_r4[OH][O]

d[H2O]/dt = -k_f2[O][H2O] + k_r2[OH][OH]

d[OH]/dt = 2(k_f2[O][H2O] - k_r2[OH][OH]) - k_f3[CO][OH] + k_r3[CO2][H] + k_f4[H][O2] - k_r4[OH][O]

d[H]/dt = k_f3[CO][OH] - k_r3[CO2][H] - k_f4[H][O2] + k_r4[OH][O]

The above equation are a system of coupled ODE's. Compact notation is used for describing ODE system of equations to computer program.

Species index (j)

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

Reaction index (i)

i	Reaction
1	CO+ O2 ⇌ CO2 + O
2	O + H2O ⇌ OH + OH
3	CO + OH ⇌ CO2 + H
4	H + O2 ⇌ OH + O

We need stoichiometric coefficients for reactants and products.

Number of rows in the matrix is equal to number of reactions.

Number of columns in the matrix is equal to number of species.

Stoichiometric coefficient matrix of reactants

Stoichiometric coefficient matrix of products

Net reaction rate (q_i) for each reaction

Where,

k_fi is the forward reaction rate

k_ri is the reverse reaction rate

[X_j] is the Concentration of species

Net reaction rate for 1^st reaction

q₁ = k_f1.[CO].[O2] - k_r1.[CO2].[O]

Net reaction rate for 2^nd reaction

q₂ = k_f2.[O].[H2O] - k_r2.[OH]²

Net reaction rate for 3^rd reaction

q₃ = k_f3.[CO].[OH] - k_r3.[CO2].[H]

Net reaction rate for 4^th reaction

q₄ = k_f4.[O2].[H] - k_r4.[O].[OH]

Net production rate for each species

where, L is the number of reactions

ω1 = v11.q1 + v12.q2 + v13.q3 + v14.q4

ω1 = -q1 - q3 = -k_f1[CO][O2] + k_r1[CO2][O] - k_f3[CO][OH] + k_r3[CO2][O] = d[CO]/dt

ω2 = v21.q1 + v22.q2 + v23.q3 + v24.q4

ω2 = -q1 - q4 = d[O2]/dt = -k_f1[CO][O2] + k_r1[CO2][O] - k_f4[H][O2] + k_r4[OH][O]

ω3 = v31.q1 + v32.q2 + v33.q3 + v34.q4

ω3 = q1 + q3 = d[CO2]/dt = k_f1[CO][O2] - k_r1[CO2][O] + k_f3[CO][OH] - k_r3[CO2][H]

ω4 = v41.q1 + v42.q2 + v43.q3 + v44.q4

ω4 = = q1 - q2 + q4 = d[O]/dt = k_f1[CO][O2] - k_r1[CO2][O] -k_f2[O][H2O] + k_r2[OH][OH] + k_f4[H][O2] - k_r4[OH][O]

ω5 = v51.q1 + v52.q2 + v53.q3 + v54.q4

ω5 = = -q2 = d[H2O]/dt = -k_f2[O][H2O] + k_r2[OH][OH]

ω6 = v61.q1 + v62.q2 + v63.q3 + v64.q4

ω6 = 2q2 - q3 + q4 = d[OH]/dt = 2(k_f2[O][H2O] - k_r2[OH][OH]) - k_f3[CO][OH] + k_r3[CO2][H] + k_f4[H][O2] - k_r4[OH][O]

ω7 = v71.q1 + v72.q2 + v73.q3 + v74.q4

ω7 = = q3 - q4 = d[H]/dt = k_f3[CO][OH] - k_r3[CO2][H] - k_f4[H][O2] + k_r4[OH][O]

From the above equation and analysis, the reaction rate of ODEs & production rate of each species for the given simple reaction mechanism are determined.