Theoretical Oxygen Demand (ThOD)

C_{6} H_{12} O_{6} + 6 O_{2} \to 6 CO_{2} + 6 H_{2} O

ThOD is the stoichiometric oxygen required to oxidize an organic compound to CO₂ and H₂O (and NH₃/HNO₃ if nitrogen is present). For glucose, 180 g compound needs 6×32 = 192 g O₂, so ThOD = 192/180 = 1.067 g O₂/g. In general ThOD ≥ COD ≥ BOD because COD misses some refractory organics’ theoretical demand only slightly, while BOD is the biodegradable fraction.

$ThOD$ (mg/L): Theoretical oxygen demand
$COD$ (mg/L): Chemical oxygen demand
$BOD$ (mg/L): Biochemical oxygen demand

Step-by-step solved example

A wastewater contains 300 mg/L glucose (C6H12O6). ThOD?

Oxygen demand of a waste stream entering a reactor.

1. Stoichiometry
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O. MW_glucose = 180; 6 O2 = 192 g.
2. Mass ratio
ThOD = 300 × (192/180) = 300 × 1.067 = 320 mg/L as O2.
$ThOD = 320 mg/L$

Answer: ThOD = 320 mg/L

10 practice questions

0/10 correct

1.ThOD of glucose in g O₂ per g glucose is

2.180 mg/L glucose has ThOD =

3.In typical municipal wastewater the inequality is

4.Ethanol C2H5OH + 3 O2 → 2 CO2 + 3 H2O. ThOD (g/g) =

5.COD uses a strong oxidant such as

6.NBOD is extra oxygen for

7.If a compound is not biodegradable, BOD5 is near 0 while ThOD is

8.Glucose ThOD factor 192/180 comes from

9.50 mg/L glucose → ThOD =

10.COD is typically run because it is