KINETICS AND STOICHIOMETRY

Stoichiometries allow expressing rates of consumption or formation of any component during reactions in terms of the other components since stoichiometric coefficients reflect the proportion in which reaction components react.

First, convert stoichiometry (with arrow, ) to equation, with = sign:

Sign convention is + for formation and – for consumption.

So for a mass-based stoichiometry, conservation of mass in a reaction with A1 as the reference compound is expressed:

(-1)A1 + (-Ψ2) A2+…+(-Ψk) Ak+…+(Ψk+1) Ak+1+…+(Ψm) Am = 0 (1)

For any reaction, there is an overall reaction rate, r, usually related to the rate of formation or consumption of the reference compound such that for reaction (1):

Example:

Given the COD stoichiometry and cell yield of 0.7 g-CODcells/g-CODS:

(-1)CODS – (1-0.7)O2 + (0.7)CODcells = 0

And r1 = rate of soluble COD consumption, r2 = rate of O2 consumption, and r3 = rate of cell formation (mg/l-hr). If cells are produced at a rate of 200 mg/l-hr = r3, what are the rates of soluble COD and oxygen consumption? Since we are using COD, substitute Y for Ψ:

r = r3/Y3 = 200 mg-CODcells/l-hr/0.7mg-CODcells/mg-CODS = 286 COD/l-hr

r1 = -Y1*r = -1*286 mg-CODS/l-hr = -286 mg-CODS/l-hr

r2 = -Y2*r = -(1-0.7)mg-O2/mg-COD*286 mg-COD/l-hr = -86 mg-O2/l-hr

NOTE, you can express oxygen as COD and then r2 is a “COD formation” rate:

r2’ = Y2*r = 86 mg-CODO2/l-hr where r2’ is the rate of COD equivalents of oxygen produced.

For the COD stoichiometry:

-1+(1-0.7)+0.7 = 0 

And the same is true for the rates when all terms are COD units:

r1 + r2 +r3 = 0

-286 + 86 + 200 (mg-COD/l-hr) = 0 

This format is especially useful when considering multiple reactions involving some or all of a set of components. For m components and n reactions:

Stoichiometry rate

(-1)A1 + (-Ψ2,1) A2+…+(-Ψk,1) Ak+…+(Ψk+1,1) Ak+1+…+(Ψm,1) Am = 0 r1

(Ψ1,2)A1 + (-Ψ2,2) A2+…+(-Ψk,2) Ak+…+() Am = 0 r2

(-Ψ1,3)A1 + (Ψ2,3) A2+…+(Ψk,3) Ak+…+(Ψk+1,3) A k+1+…+(-Ψm,3) Am = 0 r3

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(Ψ1,n)A1 + (Ψ2,n) A2+…+(-Ψk,n) Ak+…+(Ψk+1,n) Ak+1+…+(Ψm,n) Am = 0 rn

Note that the ith component, Ai, can be a product or a reactant, depending on the reaction. For example, cells are a product in the growth reaction and a reactant in decay.

Then for each reaction, mass is conserved, as with

And the net rate of formation/consumption of the ith component, ri, over all reactions, rj for j = 1  n, is:

Example:

Reaction 1 (cell growth on soluble COD):

(-1)CODS – (1-0.7)O2 + (0.7)CODcells = 0

Reaction 2 (cell CODcells decay and generation of soluble microbial products, CODS, and debris, CODdebris) given 20% of decayed cells become debris and 80% become soluble COD:

(1-0.2)CODS + (-1)CODcells + 0.2CODdebris = 0

Then

For the coupled reactions 1 and 2, rS = net rate of soluble CODS consumption, rO = net rate of soluble oxygen (as COD) consumption, rXB = net rate of cell formation, and rXD = net rate of debris formation. (all units mg-COD/l-hr)

rS = (-1)*r1 + (1-0.2)*r2

rO = (1-0.7)r1

rXB = (0.7)*r1 – 1*r2

rXD = 0.2*r2

4 equations, 6 unknowns, need two values to solve:

Given rO = 300 mg-CODO/l-hr, and rXB = 100 mg-CODcells/l-hr, find rS, rXD, r1 and r2.

Solution:

r1 = rO*(1-0.7) = 300/0.3 = 1,000 mg-COD/l-hr

r2 = -1*rXB + 0.7*r1 = -100 +700 = 600 mg-COD/l-hr

rXD = 0.2*600 = 120 mg-COD/l-hr

rS = -1000 + 0.8*600 = -520 mg-COD/l-hr