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Week 5.1 - Compact Notation Derivation for a simple Mechanism

AIM: Derive the compact Notation of simple Mechanism. Reaction Mechanism: In chemical kinetics, we use measurement of the macroscopic properties like,rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This…

Vishavjeet Singh Yadav
updated on 21 Jan 2022

AIM: Derive the compact Notation of simple Mechanism.

Reaction Mechanism:

In chemical kinetics, we use measurement of the macroscopic properties like,rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular level describes how individual atoms, ions, molecules interect to form products and all these stepwise changes are collectively called reaction mechanism.

Species Notations:

j	Species
1	$C O$ $CO$
2	$O_{2}$ $O_2$
3	$C O_{2}$ $CO_2$
4	$O$ $O$
5	$H_{2} O$ $H_2O$
6	$O H$ $OH$
7	$H$ $H$

Reaction Notation:

i	Reactions
$R_{1}$ $R_1$	$C O + O_{2} \Leftrightarrow C O_{2} + O$ $CO + O_2 iff CO_2 +O$
$R_{2}$ $R_2$	$O + H_{2} O \Leftrightarrow O H + O H$ $O + H_2O iff OH +OH$
$R_{3}$ $R_3$	$C O + O H \Leftrightarrow C O_{2} + H$ $CO + OH iff CO_2 + H$
$R_{4}$ $R_4$	$H + O_{2} \Leftrightarrow O H + O$ $H + O_2 iff OH +O$

Creating matrix for reactant side

$v_{j i}' = [\begin{matrix} 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \end{matrix}]$ $v_(ji)' = [[1,1,0,0,0,0,0],[0,0,0,1,1,0,0],[1,0,0,0,0,1,0],[0,1,0,0,0,0,1]]$

matrix for product side

$v_{j i}'' = [\begin{matrix} 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 & 0 \end{matrix}]$ $v_(ji)'' = [[0,0,1,1,0,0,0],[0,0,0,0,0,2,0],[0,0,1,0,0,0,1],[0,0,0,1,0,1,0]]$

Thus,

$v_{j i} = v_{j i}'' - v_{j i}'$ $v_(ji) = v_(ji)'' - v_(ji)'$

$v_{j i} = [\begin{matrix} - 1 - & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 1 & - 1 & 2 & 0 \\ - 1 & 0 & 1 & 0 & 0 & - 1 & 1 \\ 0 & - 1 & 0 & 1 & 0 & 1 & - 1 \end{matrix}]$ $v_(ji) = [[-1-,1,1,1,0,0,0],[0,0,0,-1,-1,2,0],[-1,0,1,0,0,-1,1],[0,-1,0,1,0,1,-1]]$

j represents species from 1 to 7 and values in matrix represents the moles of each species in reactions. Also row represents the number of reactions.

Therefor, net reaction rate can be calculated as

$q_{i} = k_{f i} \prod_{J = 1}^{N} {[X_{j}]}^{v_{j i}'} - k_{f i} \prod_{J = 1}^{N} {[X_{j}]}^{v_{j i}''}$ $q_i = k_(fi) prod_(J=1)^N [X_j] ^(v_(ji)' )-k_(fi) prod_(J=1)^N [X_j] ^(v_(ji)'' )$

Now, net reaction rate for $r_{1}$ $r_1$ is

$q_{1} = k_{f 1} [C O] [O_{2}] - k_{r 1} [C O_{2}] [O]$ $q_1 = k_(f1) [CO][O_2]-k_(r1) [CO_2][O]$

Similarly net reaction rate

$q_{2} = k_{f 2} [O] [H_{2} O] - k_{r 2} [O H] [O H]$ $q_2 = k_(f2) [O][H_2O]-k_(r2) [OH][OH]$

$q_{3} = k_{f 3} [C O] [O H] - k_{r 3} [C O_{2}] [H]$ $q_3 = k_(f3) [CO][OH]-k_(r3) [CO_2][H]$

$q_{4} = k_{f 4} [H] [O_{2}] - k_{r 4} [O H] [O]$ $q_4 = k_(f4) [H][O_2]-k_(r4) [OH][O]$

ODE's for each species

$d \frac{C O}{d t} = k_{r 1} [C O_{2}] [O] - k_{f 1} [C O] [O_{2}] + k_{r 3} [C O_{2}] [H] - k_{f 3} [C O] [O H]$ $d[CO]/dt = k_(r1)[CO_2][O]-k_(f1)[CO][O_2] +k_(r3)[CO_2][H]- k_(f3)[CO][OH]$

$d \frac{O_{2}}{d t} = k_{r 1} [C O_{2}] [O] - k_{f 1} [C O] [O_{2}] + k_{r 4} [O H] [O] - k_{f 4} [H] [O_{2}]$ $d[O_2]/dt = k_(r1)[CO_2][O]-k_(f1)[CO][O_2]+k_(r4)[OH][O]- k_(f4)[H][O_2]$

$d \frac{C O_{2}}{d t} = k_{r 1} [C O_{2}] [O] - k_{f 1} [C O] [O_{2}] + k_{r 3} [C O_{2}] [H] - k_{f 3} [C O] [O H]$ $d[CO_2]/dt = k_(r1)[CO_2][O]-k_(f1)[CO][O_2]+k_(r3)[CO_2][H]-k_(f3)[CO][OH]$

$d \frac{O}{d t} = k_{r 1} [C O_{2}] [O] - k_{f 1} [C O] [O_{2}] + k_{r 2} [O H] [O H] - k_{f 2} [O] [H_{2} O] + k_{r 4} [O] [O H] - k_{f 4} [H] [O_{2}]$ $d[O]/dt = k_(r1)[CO_2][O]-k_(f1)[CO][O_2]+k_(r2)[OH][OH]- k_(f2)[O][H_2O]+k_(r4)[O][OH]- k_(f4)[H][O_2]$

$d \frac{H_{2} O}{d t} = k_{r 2} [O H] [O H] - k_{f 2} [O] [H_{2} O]$ $d[H_2O]/dt = k_(r2)[OH][OH]- k_(f2)[O][H_2O]$

$d \frac{O H}{d t} = k_{r 2} [O H] [O H] - k_{f 2} [O] [H_{2} O] + k_{r 3} [C O_{2}] [H] - k_{f 3} [C O] [O H]$ $d[OH]/dt =k_(r2)[OH][OH]- k_(f2)[O][H_2O] +k_(r3)[CO_2][H]- k_(f3)[CO][OH]$

$d \frac{H}{d t} = k_{r 3} [C O_{2}] [H] - k_{f 3} [C O] [O H] + k_{r 4} [O H] [O] - k_{f 4} [H] [O_{2}]$ $d[H]/dt =k_(r3)[CO_2][H]- k_(f3)[CO][OH]+k_(r4)[OH][O]- k_(f4)[H][O_2]$

Product rate can be defined as:

$w_{j} = \sum_{i = 1}^{L} v_{j i} q_{i}$ $w_j = sum_(i=1)^L v_(ji)q_i$

Thus the product rate for the reactions can be written as,

For species CO

$w_{1} = v_{11} q_{1} + v_{12} q_{2} + v_{13} q_{3} + v_{14} q_{4}$ $w_1 =v_11q_1+v_12q_2+v_13q_3+v_14q_4$

$w_{1} = (- 1) q_{1} + (0) q_{2} + (- 1) q_{3} + (0) q_{4}$ $w_1 =(-1)q_1+(0)q_2+(-1)q_3+(0)q_4$

$w_{1} = - k_{f 1} [C O] [O_{2}] + k_{r 1} [C O_{2}] [O] - k_{f 3} [C O] [O H] + k_{r 3} [C O_{2}] [H]$ $w_1 = -k_(f1)[CO][O_2]+k_(r1)[CO_2][O] -k_(f3)[CO][OH]+k_(r3)[CO_2][H]$

ODE can be written as

$d \frac{w_{1}}{d t} = d \frac{C O}{d t}$ $d(w_1)/dt =d[CO]/dt$

We have seen here that the equation is same as we extracted above.

Similarly product rate can be write for other species

$w_{2} = v_{21} q_{1} + v_{2} q_{2} + v_{23} q_{3} + v_{24} q_{4}$ $w_2 =v_21q_1+v_2q_2+v_23q_3+v_24q_4$

Thus the equations can write in matrix form....

$[\begin{matrix} w_{1} \\ w_{2} \\ w_{3} \\ w_{4} \\ w_{5} \\ w_{6} \\ w_{7} \end{matrix}] = {[\begin{matrix} - 1 & - 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 1 & - 1 & 2 & 0 \\ - 1 & 0 & 1 & 0 & 0 & - 1 & 1 \\ 0 & - 1 & 0 & 1 & 0 & 1 & - 1 \end{matrix}]}^{T} X [\begin{matrix} q_{1} \\ q_{2} \\ q_{3} \\ q_{4} \end{matrix}]$ $[[w_1],[w_2],[w_3],[w_4],[w_5],[w_6],[w_7]] = [[-1,-1,1,1,0,0,0],[0,0,0,-1,-1,2,0],[-1,0,1,0,0,-1,1],[0,-1,0,1,0,1,-1]]^T X[[q_1],[q_2],[q_3],[q_4]]$