Potential Carbonaceous BOD of 1 g of Cells: 18-Env-B9 Environmental Chemistry and Microbiology, May 2013, Q21

Question 21: Potential Carbonaceous BOD of 1 g of Cells (4 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Same empirical biomass formula as Chemistry Question 11, $\text{C}_5\text{H}_7\text{NO}_2$, atomic weights C=12, H=1, N=14, O=16.

Find. The ultimate carbonaceous BOD ($\text{BOD}_u$, oxygen equivalent) of 1 g of cell mass.

Approach. The potential (ultimate) carbonaceous BOD of a compound equals its theoretical COD, since both represent the total oxygen required for complete biological oxidation of the carbonaceous material (with organic-N terminating at $\text{NH}_3$, not carried through to nitrogenous demand) — the same balanced reaction and molecular-weight ratio computed in Chemistry Q11 applies directly.

Reuse the balanced oxidation and molar mass from Q11 (Chemistry). $$\text{C}_5\text{H}_7\text{NO}_2 + 5\text{O}_2 \longrightarrow 5\text{CO}_2+2\text{H}_2\text{O}+\text{NH}_3, \qquad MW_{\text{cells}}=113\ \text{g/mol}$$
Oxygen equivalent per gram of cells. $$\text{BOD}_u = \frac{5\times32}{113} = 1.416\ \text{g O}_2\text{/g cells}$$ For $1\ \text{g}$ of cells, $$\boxed{\text{BOD}_u(1\ \text{g cells}) = 1.42\ \text{g O}_2}$$

Potential carbonaceous BOD — final results
Quantity	Value
Ultimate carbonaceous BOD per gram of cells	1.42 g O₂/g
BOD₊ of 1 g of cells	1.42 g O₂

Concept & Theory

Topic. Ultimate carbonaceous BOD ($\text{BOD}_u$) of biomass, and its relation to theoretical COD.

Key relations. $\text{BOD}_u \approx \text{COD}$ for a fully biodegradable organic compound; $\text{BOD}_u = \text{COD}\times(1-e^{-k\cdot\infty})=\text{COD}$ in the limit of complete oxidation.

Why this works. COD measures the total chemical oxidizing capacity required, while BOD measures the biologically exerted portion; for a readily biodegradable substrate like bacterial cell material, the ultimate (long-time) BOD converges to the same theoretical oxygen demand as COD, because essentially all of the organic carbon is eventually biologically oxidizable.

Common pitfall. Reporting the standard 5-day BOD ($\text{BOD}_5$) instead of the ultimate/potential BOD the question asks for — $\text{BOD}_5$ is only a fraction of $\text{BOD}_u$, governed by the rate constant $k$ (typically $\text{BOD}_5/\text{BOD}_u\approx0.68$ for domestic wastewater at $k=0.23\ \text{d}^{-1}$), and is not what is computed here.

Source. Metcalf & Eddy, ch. 8 (BOD/COD relationships, endogenous decay).

Question 21 of 25: Potential Carbonaceous BOD of 1 g of Cells

Section 1: Chemistry (11 questions, 50 marks)

Question 21: Potential Carbonaceous BOD of 1 g of Cells (4 marks)