Compact Notation Derivation for a simple Mechanism

Objective:

To derive the reaction rate Ordinary Differential Equations (ODE) for the reaction mechanism below,

`CO + O_2 rightarrow CO_2 + O`

`O + H_2O rightarrow OH + OH`

`CO +OH rightarrow CO_2 + H`

`H + O_2 rightarrow OH + O`

Theory:

A simple chemical reaction contains several elementary intermediate reactions, which occur for a definite time period. The set of all the intermediate reactions, including the initial and final reactions is called as a reaction mechanism.

The intermediate reactions are dependent on several parameters like initial conditions, time and so on. Using these parameters, by controlling them, the end result can be predicted.

The time rate of change of each species is calculated using rate law, which states that the time rate of change of the species of a reaction is directly proportional to the concentration or quantity available at that instant. Using this law the reaction rates of the species involved in the reactions are found out.

`(d[X])/dt prop [X]`

When the proportionality is removed, a constant is multiplied to the RHS (linearly proportional),

`(d[X])/dt = K_(forr) * [X]` where `K` is rate of forward or reaction reaction.

In this case, as the reaction are all forward, `K_r` would be zero.

To derive the ODEs required to simulate the combustion mechanism, com

Introduction:

The data required to perform simulation is fed into the machine in the form of matrices, so the ODE data needs to be framed into matrices.

The first step is to generate a matrix with number of moles of each species in each reaction for both reactants and the products. Number of species involved are seven and the number of reactions are four, so the size of stoichiometric coefficient matrices would be 4x7.

Then, the difference between them is found out (products matrix minus reactants matrix). Then, the net reaction rate is found out for each reaction, the sign of which denotes the direction in which the reaction is taking place, forward or backward. After calculating the net reaction rate, net production rate of each species is calculated.

Formulation:

Step 1: Compact notation

Species	j
`CO`	1
`O_2`	2
`CO_2`	3
`O`	4
`H_2O`	5
`OH`	6
`H`	7

Reaction	i
Reaction 1	1
Reaction 2	2
Reaction 3	3
Reaction 4	4

Step 2: Matrix formation

Matrix denoting number of moles of each species in each reaction in reactants side,

`v_(ji)^I` (Number of moles of species i in reaction j) = `[[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]]`

Matrix denoting number of moles of each species in each reaction in products side,

`v_(ji)^(II)` = `[[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]]`

Now, `v_(ji) = v_(ji)^(II) - v_(ji)^I`, which gives,

`v_(ji) = [[-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]]`

Step 3: Net reaction rate

Net reaction rate is calculated from the expression,

`q_i = K_(fi)*prod_(j=1)^N [X_j]^(v_(ji)^I) - K_(ri)*prod_(j=1)^N [X_j]^(v_(ji)^(II))`

Note that, all reactions in this case are irreversible and forward, so `K_(ri) = 0`

After substituting the corresponding values in the above expression, for four reactions, we get four net reaction rates which are,

`q_1 = K_(f1)*[CO][O_2]`

`q_2 = K_(f2)*[O][H_2O]`

`q_3 = K_(f3)*[CO][OH]`

`q_4 = K_(f4)*[H][O_2]`

Step 4: Net production rate

Net production rate is calculated for each species, which is calculated using the expression,

`omega_j = sum_(i=1)^N v_(ji)*q_i`

After calculating net production rate for each species, it is formulated in the matrix form as,

`[[omega_1],[omega_2],[omega_3],[omega_4],[omega_5],[omega_6],[omega_7]]` = `[v_(ji)]^T* [[q_1],[q_2],[q_3],[q_4]]`

NOTE: Transpose of `v_(ji)` matrix is taken to ensure matrix multiplication is possible

From the above expression, net production rate of each species can be calculated.

Conclusion:

The net reaction rate and net production rate of each reaction and each species is calculated respectively. And the data is formulated in matrices form, which can be fed to a machine for simulation.