Week 5.1 - Compact Notation Derivation for a simple Mechanism

AIM: The aim of this challenge is to derive the reaction rate and the production rate for the given simple mechanism or set of reactions.

OBJECTIVES:

The objectives of the challenge is to identify different species in the reaction.

Generate the matrices for reactants, products and for net reaction rate.

THEORY:

The given reaction Mechansim is :

`CO + O_2-> CO_2+O`

`O+H_2O-> OH+OH`

`CO+OH->CO_2+H`

`H+O_2->OH+O`

For this mechanism the reaction rate for reactants nad products are denoted by `k_f and k_r` respectively.

The positive and negative signs for are decided based on the whether the species is gnerating or delpeting.

`d [ C O ] d t = − 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 ]`

`d [ O 2 ] d t = − k f 1 [ C O ] [ O 2 ] + k r 1 [ C O 2 ] [ O ] − k f 4 [ H ] [ O 2 ] + k r 4 [ O H ] [ O ]`

`d [ C O 2 ] d t = 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 ]`

`d [ O ] d t = k f 1 [ C O ] [ O 2 ] − k r 1 [ C O 2 ] [ O ] − k f 2 [ O ] [ H 2 O ] + k r 2 [ O H ] [ O H ] + k f 4 [ H ] [ O 2 ] − k r 4 [ O H ] [ O ]`

`d [ H 2 O ] d t = − k f 2 [ O ] [ H 2 O ] + k r 2 [ O H ] [ O H ]`

`d [ O H ] d t = 2 ( k f 2 [ O ] [ H 2 O ] − k r 2 [ O H ] [ O H ] ) − k f 3 [ C O ] [ O H ] + k r 3 [ C O 2 ] [ H ] + k f 4 [ H ] [ O 2 ] − k r 4 [ O H ]`

`d [ H ] d t = k f 3 [ C O ] [ O H ] − k r 3 [ C O 2 ] [ H ] − k f 4 [ H ] [ O 2 ] + k r 4 [ O H ] [ O ]`

Compact notation for the programming puprose:

The compact notation starts by defining variables. Species index is denoted by `'j'` and the reaction number is denoted by `'i'`.

Species `CO` `O_2` `CO_2` `O` `H_2O` `OH` `H`

INDEX(`j`) 1 2 3 4 5 6 7

Reactions iNDEX(`i`)

`CO + O_2-> CO_2+O` 1

`O+H_2O-> OH+OH` 2

`CO+OH->CO_2+H` 3

`H+O_2->OH+O` 4

To define the net reaction rate matrix which is `v_(ji) = v_(ji)^('')-v_(ji)^'` where `v_(ji)^('')` is for products and `v_(ji)^(')` is for reactants.

The reactant matrix can be written as :

`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]]`

The product matrix can be written as :

`v_(ji)^('')` =`[[0,1,1,0,0,0,0],[0,0,0,0,0,2,0],[0,0,1,0,0,0,1],[0,0,0,1,0,1,0]]`

Hence the net reaction matrix is :

`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]]`

The net reaction rate for each state is given by forward minus reverse reacton:

`q _i = k _(f i) ∏_( j = 1)^N [ X _J ]^( v j i ') − k _(r i) ∏_ (j =1)^ N [ X_ J ] ^(v j i ^(' '))`

Hence for `j` = 1 to 4 we have :

`1 . q_1 = k _(f 1) [ C O ]^ 1 [ O _2 ]^ 1 [ C O_ 2 ]^ 0 [ O ]^ 0 [ H_ 2 O ]^ 0 [ O H ]^ 0 [ H ]^ 0 − k_ (r 1) [ C O ]^ 0 [ O_2 ] ^0 [ C O_2 ]^ 1 [ O ]^1 [ H_ 2 O ]^ 0 [ O H ]^0 [ H ]^ 0`

`q_1 = k_(f 1) [ C O ]^ 1 [ O_ 2 ]^ 1 − k (r 1) [ C O_ 2 ]^ 1 [ O ]^ 1`

`2 . q_ 2 = k_( f 2) [ C O ]^ 0 [ O_2 ]^0 [ C O_2 ]^ 0 [ O ]^ 1 [ H_ 2 O ]^ 1 [ O H ]^ 0 [ H ]^ 0 − k_ (r 2) [ C O ]^ 0 [ O_ 2 ] ^0 [ C O_2 ]^ 0 [ O ]^ 0 [ H_2 O ]^ 0 [ O H ]^ 2 [ H ] ^0`

`q_2 = k_(f 2) [ O ]^ 1 [ H_ 2 O ]^ 1 − k_(r 2) [ O H ]^ 2`

`3 . q_3 = k_( f 3) [ C O ]^ 1 [ O_ 2 ]^ 0 [ C O_ 2 ]^ 0 [ O ]^ 0 [ H_ 2 O ]^0 [ O H ]^1 [ H ]^ 0 − k_(r 3) [ C O ]^ 0 [ O_ 2 ]^ 0 [ C O_2 ]^ 1 [ O ]^0 [ H_2 O ]^ 0 [ O H ]^0 [ H ] ^1`

`q_3 = k_(f 3) [ C O ]^ 1 [ O H ]^1 − k_ (r 3) [ C O_2 ]^ 1 [ H ]^ 1`

`4 . q_ 4 = k_( f 4) [ C O ]^ 0 [ O_ 2 ]^1 [ C O_2 ]^ 0 [ O ]^ 0 [ H_ 2 O ]^0 [ O H ]^0 [ H ]^ 1 − k_( r 4)[ C O ]^ 0 [ O_ 2 ]^ 0 [ C O_2 ]^ 0 [ O ]^1 [ H_ 2 O ]^0 [ O H ]^ 1 [ H ]^ 0`

`q _4 = k _(f 4) [ O_2 ]^1 [ H ]^1 − k_( r 4) [ O ] ^1 [ O H ]^ 1`

Now `w_ j = sum_(i=1)^L = v_( j i) ^t q _i`

where `w`= net production of series

`([w_1],[w_2],[w_3],[w_4],[w_5],[w_6])= [[-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]][[q_1],[q_2],[q_3],[q_4]]`

Evaluating these values and also that `w_ j = (d [ X_ j ])/ (d t)`where `L`= no. of reactions

`1 . w_1 = (d [ C O ]) /(d t) = sum_1^L v _(1 i) q_ i = v_( 11) q _1 + v_ (12) q_ 2 + v_( 13) q_ 3 + v_ (14) q_ 4`

Applying values from matrix

`w _1 = − q_ 1 − q_ 3`

`w_1 = (d [ C O ])/ (d t) = − ( k_ (f 1) [ C O ]^ 1 [ O_ 2 ]^ 1 − k _(r 1) [ C O_ 2 ]^ 1 [ O ]^ 1 ) − ( k_ (f 3) [ C O ]^ 1 [ O H ]^ 1 − k _(r 3) [ C O _2 ]^ 1 [ H ]^ 1 )`

`2 . w_ 2 = (d [ O 2 ])/ (d t) = sum_1^L v_(2 i) q_ i = v_( 21) q_1 + v _(22) q _2 + v_(23) q_3 + v_ (24) q_ 4`

Applying values from matrix

`w _2 = − q_ 1 − q_ 4`

`w_ 2 = (d [ O 2 ])/ (d t) = − ( k_ (f 1) [ C O ]^ 1 [ O_2 ]^ 1 − k_ (r 1) [ C O_ 2 ]^ 1 [ O ]^ 1 ) − ( k _(f 4) [ O_2 ]^ 1 [ H ]^ 1 − k_ (r 4) [ O ]^ 1 [ O H ]^ 1 )`

`3 . w_ 3 = (d [ C O 2 ])/ (d t) = sum_1^L v_ (3 i) q_ i = v_ (31) q_1 + v_ (32) q_ 2 + v_(33) q_ 3 + v _(34) q _4`

Applying values from matrix

`w_ 3 = q _1 + q_ 3`

`w_ 3 = (d [ C O 2 ])/ (d t) = ( k_( f 1) [ C O ]^ 1 [ O_ 2 ]^ 1 − k_ (r 1) [ C O_ 2 ]^ 1 [ O ]^ 1 ) + ( k_ (f 3) [ C O ]^ 1 [ O H ]^ 1 − k _(r 3) [ C O_ 2 ]^ 1 [ H ]^ 1 )`

`4 . w_ 4 = (d [ O ])/ (d t) = sum_1^L v_ (4 i) q_i = v_ (41) q_ 1 + v_( 42) q_2 + v_ (43) q_ 3 + v _(44) q _4`

Applying values from matrix

`w _4 = q _1 − q_ 2 + q _4`

`w_ 4 = (d [ O ])/ (d t) = ( k_ (f 1) [ C O ]^ 1 [ O_ 2 ]^ 1 − k_ (r 1) [ C O_2 ]^ 1 [ O ]^ 1 ) − ( k_ (f 2) [ O ]^ 1 [ H_ 2 O ]^ 1 − k_( r 2) [ O H ]^ 2 ) + ( k_ (f 4) [ O 2 ]^ 1 [ H ]^ 1 − k _(r 4) [ O ]^ 1 [ O H ]^ 1 )`

`5 . w _5 = (d [ H 2 O ])/ (d t) = sum_1^Lv_(5 i) q_i = v_ (51) q_ 1 + v _(52) q_2 + v_(53) q_ 3 + v _(54) q_ 4`

Applying values from matrix

`w_ 5 = − q_ 2`

`w_ 5 = (d [ H 2 O ])/ (d t) = − ( k _(f 2) [ O ]^ 1 [ H_ 2 O ]^ 1 − k_( r 2) [ O H ]^ 2 )`

`6 . w_6 = (d [ O H ])/ (d t) = sum_1^L v_(6 i) q_i = v_61 q_1 + v_62 q_2 + v_63 q_3 + v_64 q_4`

Applying values from matrix

`w_ 6 = 2 q_2 − q_3 + q_4`

`w_6 = (d [ O H ])/ (d t) = 2 ( k_( f 2) [ O ]^ 1 [ H_ 2 O ]^ 1 − k _(r 2) [ O H ]^ 2 ) − ( k_ (f 3) [ C O ]^ 1 [ O H ]^ 1 − k _(r 3) [ C O_ 2 ]^ 1 [ H ]^ 1 ) + ( k_ (f 4) [ O_2 ]^ 1)`

`7 . w_7 = (d [ H ])/ (d t) = sum_1^L v_(7 i) q_i = v_( 71) q_1 + v_72 q_2 + v_73 q_3 + v_74 q_4`

Applying values from matrix

`w_7 = q_3 − q_4`

`w_7 = (d [ H ])/(d t) = ( k_(f 3) [ C O ]^ 1 [ O H ]^ 1 − k_(r 3) [ C O_2 ]^ 1 [ H ]^ 1 ) − ( k_(f 4) [ O_2 ]^ 1 [ H ]^ 1 − k_(r 4) [ O ]^ 1 [ O H ]^ 1 )`

In this way we predict the ODE for the reaction rate and production rate of given mechanism