Question

Anaerobic bacteria are being utilized for the destruction of biodegradable organic content of wastewater measured in terms of Chemical Oxygen Demand (COD). The yield coefficient (Y) is 0.06 gram Volatile Suspended Solids (VSS) per gram of COD. Bacterial biomass forms the sludge during the process. The proximate chemical formula of anaerobic bacteria is \(C_6H_7NO_2\).

The COD (in %) converted into bacterial biomass is (rounded off to the nearest integer).

Hint:

Find the oxygen (COD) equivalent of one mole of the cell formula \(C_6H_7NO_2\) by balancing its oxidation reaction, then multiply that by the yield coefficient to get the fraction of feed COD that ends up in the biomass.

Answer 1

Correct Answer: 9

Step 1: Set up the electron-equivalent idea.
Instead of writing out the full balanced oxidation reaction, we can find the oxygen demand of the bacterial cells using electron equivalents. Every mole of an organic compound that gets fully oxidised gives up a fixed number of electrons, and each electron equivalent corresponds to a fixed mass of oxygen.

Step 2: Count the electron equivalents in the cell formula.
For an organic compound $C_xH_yO_zN_w$ that is oxidised aerobically to $CO_2$, $H_2O$ and $NH_3$, the number of electron equivalents per mole works out to:
\[ n = 4x + y - 2z - 3w \] Here $x=6$, $y=7$, $z=2$, $w=1$, taken from $C_6H_7NO_2$, so:
\[ n = 4(6) + 7 - 2(2) - 3(1) = 24 + 7 - 4 - 3 = 24 \text{ equivalents per mole} \]
Step 3: Convert electron equivalents into an oxygen mass.
One electron equivalent of oxygen weighs 8 g, since $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$ shows 4 equivalents per mole of $O_2$, and $32/4 = 8$. So the oxygen demand of one mole of cells is:
\[ 24 \times 8 = 192 \text{ g } O_2 \text{ per mole of cells} \] The molar mass of $C_6H_7NO_2$ is $6(12) + 7(1) + 14 + 2(16) = 125$ g/mol, so the COD-to-VSS ratio is $192/125 = 1.536$ g COD per g VSS, the same number the reaction-balance route gives, confirming the result.

Step 4: Apply the yield coefficient.
The yield $Y = 0.06$ g VSS forms per g COD removed, so for every gram of COD destroyed, $0.06$ g of new cells form. The COD locked inside those cells is:
\[ 0.06 \times 1.536 = 0.09216 \text{ g COD per g COD removed} \]
Step 5: State the final percentage.
Multiplying by 100 gives $9.216\%$, which rounds off to 9%.
\[ \boxed{9\%} \]