Question

Assume COD entering an anaerobic digester can be modeled as 2C3H5O3- = C3H5O2- + C2H3O2- + CO2 + H2 (Balanced reaction

Next, assume that methane in the reactor results from two pathways: aceticlastic methanogenesis and hydrogenotrophic methanogenesis. If we assume that all necessary carbon and hydrogen for methane production comes only from fermentation of lactate; how much methane (in mg/L) can be produced assuming 25,000mg/L COD entering the anaerobic digester, with all of it being fermented?

Answer 1

We will first write the given balanced equation in protonated form-

       2C3H6O3 = C3H6O2 + C2H4O2 + CO2 + H2

OR C3H6O3 =   1/2C3H6O2 + 1/2C2H4O2 + 1/2 CO2 +   1/2 H2…….(1)

    (mol wt = 90)                    (mol wt = 60)                       (mol wt = 2)

We will now write the acetoclastic methanogenesis equation –

        C2H4O2     =     CH4      +   CO2……..(2)

(mol wt = 60)     (mol wt = 16)    

We will also write the hydrogenotropic methanogenesis equation –

        CO2 + 4H2 = CH4 + 2H2O

OR 1/4CO2   +    H2       =         1/4CH4    +    1/2H2O……….(3)

                 (mol wt = 2)          (mol wt = 16)    

Now you see your question reduces to simple arithmetic problem where eqn (1) gives you reactants for methanogenesis and eqns (2) and (3) give you your product methane.

From eqn (1) we know 90mg of lactate (lactic acid) gives ½(60) = 30mg of acetate (acetic acid), therefore, amount of acetate produced by fermentation of 25000mg/L of COD will be given by

30 x 25000/90 = 8333 mg/L of acetate

From eqn (2) we know 60 mg of acetate produce 16 mg of methane, therefore, methane produced during acetoclastic methanogenesis is given by

16 x 8333/60 = 2222 mg/L of methane …..(A)

In hydrogenotropic methanogenesis hydrogen is limiting substrate i.e. amount of methane produced will depend on available amount of hydrogen.

Now from equation (1), we know 90mg of lactate (lactic acid) gives ½(2) = 1mg of hydrogen. Therefore, amount of hydrogen produced by fermentation of 25000mg/L of COD will be given by

1 x 25000/90 = 278 mg/L of hydrogen

From eqn (3) we know 2 mg of hydrogen produce ¼(16 mg) = 4mg of methane, therefore, methane produced during hydrogenotropic methanogenesis is given by

4 x 278/2 = 556 mg/L of methane …..(B).

Therefore, total amount of methane produced will be (A) + (B) = 2222 + 556 = 2778 mg/L of methane.

Thanks!