Week 5.1 - Compact Notation Derivation for a simple Mechanism
AIM - To apply compact notation derivation to a simle reaction mechanism.
OBJECTIVE - To find the net reaction rete for all the foru reaction using the reactant and product matrix and then find the production rate of each species using the matrix notation wrt the net reaction rates.
THEORY -
We will be using the compact notation system for a simple reaction mechanism. the four reaction mechanisms that are used here are given below:
`CO + O_2<=>CO_2+O`
`O + H_2 O <=> OH + OH`
`CO + OH <=> CO_2+ H`
`H + O_2<=>OH +O`
now here we have the following species that are denoted using a variable 'j' as shown in table
| Species | Notation(j) |
| `CO` | 1 |
| `O_2` | 2 |
| `CO_2` | 3 |
| `O` | 4` |
| `H_2O` | 5 |
| `OH` | 6 |
| `H` | 7 |
Now similarly we willbe using the notation variable 'i' to denote the reaction so in the following table we have the reactions and the corredponding value of the notation 'i'
| Reaction | Notation(i) |
| `CO + O_2<=>CO_2+O` | 1 |
| `O + H_2 O <=> OH + OH` | 2 |
| `CO + OH <=> CO_2+ H` | 3 |
| `H + O_2<=>OH +O` | 4 |
Now we will be creating a matrix that can store the information for each reaction and the species in that reaction .
To create this matrix we will be first creating a matrix( `v_(ji)'`) that stores the infromation for the stoichiometric coefficient of all the reactant of each reaction`
here in the name 'j' denotes the species index and 'i' denotes the reaction index, the single prime notation is used to represent the reactants.
so now the reactant matrix `v_(ji)'` will be as shown below
`v_(ji)' = [[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]]`
Similarly in this way we will be creating the product matrix that store the information for the stoichiometric coefficients of the the products for all the reactions
the product matrix is denoted by `v_(ji)''` here double prime is used to denote the products and the j and i represents the species and reaction respectively.
so now the product matrix `v_(ji)''`will be as shown below
`v_(ji)'' = [[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]]`
`v_(ji)`is the difference between the product side stiochiometric matrix and the reactant side stiochiomatric matrix, it can be given as
`v_(ji) = v_(ji)'' - v_(ji)'`
`therefore 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]]`
now we will be dreiving the net reaction rate using the formula
`q_i = k_(fi) underset(j=1)(prod^n) [X_j]^(v_(ji)') - k_(ri) underset(j=1)(prod^n) [X_j]^(v_(ji)'')`
here we can see that the first term has `k_(fi)`which is the forward rate of reaction for the i-th reaction and similarly in the sescornd term we have `k_(ri)`which the reverse rate of reaction for the i-th reaction.
`[X_j]`stands for the concentration of the j-th species and n is the total number of species ,here for this reaction mechanism n = 7.
so using the compact notation we will now expand our net reaction rate equations for each reaction
`q_1 = k_(f1) underset(j=1)(prod^n) [X_j]^(v_(j1)') - k_(r1) underset(j=1)(prod^n) [X_j]^(v_(j1)'')`
`q_1 = k_(f1). [X_1]^(v_11').[X_2]^(v_21').[X_3]^(v_31').[X_4]^(v_41').[X_5]^(v_51').[X_6]^(v_61')[X_7]^(v_71') - k_(r1) .[X_1]^(v_11'').[X_2]^(v_21'').[X_3]^(v_31'').[X_4]^(v_41'').[X_5]^(v_51'').[X_6]^(v_61'')[X_7]^(v_71'')`
Now we will be substituting the species and the values of `v_(ji)' and v_(ji)''`
`q_1 = k_(f1). [CO]^(1).[O_2]^(1).[CO_2]^(0).[O]^(0).[H_2O]^(0).[OH]^(0)[H]^(0) - k_(r1). [CO]^(0).[O_2]^(0).[CO_2]^(1).[O]^(1).[H_2O]^(0).[OH]^(0)[H]^(0)`
`q_1 = k_(f1). [CO]^(1).[O_2]^(1) - k_(r1). [CO_2]^(1).[O]^(1)`
Similarly we can write the net reaction rates for other rections as well
`q_2 = k_(f2). [CO]^(0).[O_2]^(0).[CO_2]^(0).[O]^(1).[H_2O]^(1).[OH]^(0)[H]^(0) - k_(r2). [CO]^(0).[O_2]^(0).[CO_2]^(0).[O]^(0).[H_2O]^(0).[OH]^(2)[H]^(0)`
`q_2 = k_(f2). [O]^(1).[H_2O]^(1) - k_(r2). [OH]^(2)`
`q_3 = k_(f3). [CO]^(1).[O_2]^(0).[CO_2]^(0).[O]^(0).[H_2O]^(0).[OH]^(1)[H]^(0) - k_(r3). [CO]^(0).[O_2]^(0).[CO_2]^(1).[O]^(0).[H_2O]^(0).[OH]^(0)[H]^(1)`
`q_3 = k_(f3). [CO]^(1).[OH]^(1) - k_(r3). [CO_2]^(1).[H]^(1)`
`q_4 = k_(f4). [CO]^(0).[O_2]^(1).[CO_2]^(0).[O]^(0).[H_2O]^(0).[OH]^(0)[H]^(1) - k_(r4). [CO]^(0).[O_2]^(0).[CO_2]^(0).[O]^(1).[H_2O]^(0).[OH]^(1)[H]^(0)`
`q_4 = k_(f4). [H]^(1).[O_2]^(1) - k_(r4). [OH]^(1).[O]^(1)`
Now we have the net reaction rates for all the four reactions.
Using this reaction rates we will be calculating the production rates of each species. here we can see that the speies O2 is present in reaction 1 as well as reaction 4 so we will have to use a system of couples ODE to find the reaction rates.
we can calculate the production rate for each species usign the formula
`omega_(j) = underset(i=1)(sum^l) v_(ji)q_i`
where 'l' is the total number of reactions so for our reaction mechanism the value is 'l=4'
the coupled ODEs can be denoted using the formula
`omega_j = (d[X_j])/dt`
so for j =1 we have species CO so the production rate of CO becomes
`omega_(1) = (d[X_1])/dt = underset(i=1)(sum^l) v_(1i)q_i`
`omega_(1) = (d[CO])/dt = v_(11)q_1 +v_(12)q_2 +v_(13)q_3 +v_(14)q_4`
`omega_(1) = (d[CO])/dt = -1q_1 + 0.q_2 -1.q_3 +0.q_4`
`omega_(1) = (d[CO])/dt = -1q_1 -1.q_3`
`omega_(1) = (d[CO])/dt = -1.( k_(f1). [CO]^(1).[O_2]^(1) - k_(r1). [CO_2]^(1).[O]^(1)) -1.(k_(f3). [CO]^(1).[OH]^(1) - k_(r3). [CO_2]^(1).[H]^(1))`
`omega_(1) = (d[CO])/dt = -1.( k_(f1). [CO]^(1).[O_2]^(1) - k_(r1). [CO_2]^(1).[O]^(1)) -1.(k_(f3). [CO]^(1).[OH]^(1) - k_(r3). [CO_2]^(1).[H]^(1))`
similarly in this way the production rates of each species can be found
`omega_(2) = (d[O_2])/dt = -1.( k_(f1). [CO]^(1).[O_2]^(1) - k_(r1). [CO_2]^(1).[O]^(1)) -1.(k_(f4). [H]^(1).[O_2]^(1) - k_(r4). [OH]^(1).[O]^(1))`
`omega_(3) = (d[CO_2])/dt = 1.( k_(f1). [CO]^(1).[O_2]^(1) - k_(r1). [CO_2]^(1).[O]^(1)) +1.(k_(f3). [CO]^(1).[OH]^(1) - k_(r3). [CO_2]^(1).[H]^(1))`
`omega_(4) = (d[O])/dt = 1.( k_(f1). [CO]^(1).[O_2]^(1) - k_(r1). [CO_2]^(1).[O]^(1)) -1.(k_(f2). [O]^(1).[H_2O]^(1) - k_(r2). [OH]^(2)) +1.(k_(f4). [H]^(1).[O_2]^(1) - k_(r4). [OH]^(1).[O]^(1))`
`omega_(5) = (d[H_2O])/dt = -1.(k_(f2). [O]^(1).[H_2O]^(1) - k_(r2). [OH]^(2))`
`omega_(6) = (d[OH])/dt = 2.(k_(f2). [O]^(1).[H_2O]^(1) - k_(r2). [OH]^(2)) -1.(k_(f3). [CO]^(1).[OH]^(1) - k_(r3). [CO_2]^(1).[H]^(1)) +1.(k_(f4). [H]^(1).[O_2]^(1) - k_(r4). [OH]^(1).[O]^(1))`
`omega_(7) = (d[H])/dt = 1.( k_(f3). [CO]^(1).[OH]^(1) - k_(r3). [CO_2]^(1).[H]^(1)) -1.(k_(f4). [H]^(1).[O_2]^(1) - k_(r4). [OH]^(1).[O]^(1))`
Now this system of coupled ODEs will be solved in order to get the concentration of each species.
in the matrix form it can be given as
`[[omega_1],[omega_2],[.],[.],[omega_n]] = [[(d[X_1])/dt],[(d[X_2])/dt],[.],[.],[(d[X_n])/dt]]`
this ODEs can now be solved using any solver package for example ode45 that is used by matlab or the commercial packages like fluent have their own in built solvers.
CONCLUSION -
1. We have learned how to write the compact notation for reaction mechanism and what each notation means .
2. We have also learned that the differential equation formed in this case are couples ODEs that can be solved using any of teh solver package.