Question 27 of 28: COD and TOC of an Ethylene Glycol Wastewater

Question 27: COD and TOC of an Ethylene Glycol Wastewater (3 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 = 100 mg/L; formula C₂H₆O₂; atomic weights C = 12, H = 1, O = 16.

Find. The theoretical chemical oxygen demand (COD) and total organic carbon (TOC) of the 100 mg/L solution.

Approach. Write the balanced complete-oxidation reaction to find the stoichiometric O₂ demand per mole of glycol (this gives COD, since glycol is fully oxidizable and its COD equals its theoretical oxygen demand), and separately compute the carbon mass fraction of the molecule to get TOC.

1. Molecular weight of ethylene glycol. $$MW = 2(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62\ \text{g/mol}$$
2. Balance the complete oxidation reaction. Ethylene glycol oxidizes fully to carbon dioxide and water: $$\text{C}_2\text{H}_6\text{O}_2 + 2.5\,\text{O}_2 \longrightarrow 2\,\text{CO}_2 + 3\,\text{H}_2\text{O}$$ Checking the oxygen balance confirms the coefficient: left side $2 + 2(2.5) = 7$ oxygen atoms; right side $2(2) + 3(1) = 7$ oxygen atoms — balanced, so $2.5$ mol O₂ is consumed per mole of glycol oxidized.
3. Compute the theoretical oxygen demand (= COD) per gram of glycol. $$\text{ThOD} = \frac{2.5 \times 32}{62} = \frac{80}{62} = 1.290\ \text{g O}_2/\text{g glycol}$$
4. Scale to the given concentration. $$\text{COD} = 100\ \text{mg/L} \times 1.290 = \boxed{129.0\ \text{mg/L as O}_2}$$
5. Compute the carbon mass fraction of the molecule. There are 2 carbon atoms per molecule, so $$f_C = \frac{2 \times 12}{62} = \frac{24}{62} = 0.3871$$
6. Scale to the given concentration to get TOC. $$\text{TOC} = 100\ \text{mg/L} \times 0.3871 = \boxed{38.7\ \text{mg/L as C}}$$