Question 7 of 19: COD and TOC of an Ethylene Glycol Wastewater

Question 7: COD and TOC of an Ethylene Glycol Wastewater (5 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. Ethylene glycol concentration and its molecular formula, with atomic weights for C, H and O.

Find. The theoretical Chemical Oxygen Demand (COD) and Total Organic Carbon (TOC) of the wastewater.

Approach. Write the balanced complete-oxidation reaction to find the stoichiometric oxygen demand per mole of glycol, then scale both the oxygen demand and the carbon mass by the given concentration and molecular weight.

1. Molecular weight of ethylene glycol. $$MW = 2(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62\ \text{g/mol}$$
2. Balanced complete-oxidation reaction. For $C_aH_bO_c$, complete oxidation to CO₂ and H₂O requires $\left(a + \tfrac{b}{4} - \tfrac{c}{2}\right)$ mol O₂ per mol of compound: $$C_2H_6O_2 + \left(2 + \tfrac{6}{4} - \tfrac{2}{2}\right)O_2 \rightarrow 2CO_2 + 3H_2O$$ $$C_2H_6O_2 + 2.5\,O_2 \rightarrow 2CO_2 + 3H_2O$$
3. Theoretical COD. Each mole of glycol (62 g) demands 2.5 mol O₂ (2.5 × 32 g = 80 g O₂): $$COD = 150\ \text{mg/L} \times \dfrac{80\ \text{g O}_2/\text{mol}}{62\ \text{g/mol}}$$ $$\boxed{COD \approx 193.5\ \text{mg/L as O}_2}$$ Because ethylene glycol is fully biodegradable and non-toxic to the dichromate reflux, the theoretical COD computed here is a close estimate of the measured (5220-method) COD.
4. TOC. Each mole of glycol contains 2 mol (24 g) of carbon: $$TOC = 150\ \text{mg/L} \times \dfrac{24\ \text{g C/mol}}{62\ \text{g/mol}}$$ $$\boxed{TOC \approx 58.1\ \text{mg/L as C}}$$