I took reaction mixture to study the COD, and I got some results based on that I got COD value decreases with increasing time. But I need conclusion what is happening in the reaction mixture so COD value decreases?
In environmental chemistry, the chemical oxygen demand (COD) test is commonly used to indirectly measure the amount of organic compounds in water. Most applications of COD determine the amount of organic pollutants found in surface water (e.g. lakes and rivers) or wastewater, making COD a useful measure of water quality. It is expressed in milligrams per liter (mg/L) also referred to as ppm (parts per million), which indicates the mass of oxygen consumed per liter of solution.
Girish,
in your research COD decreases with increasing time because the bacteria in the water will grow and degrade the organic matters in your water sample.
1. in your stock solution or reaction mixture you have some volatile organic compounds, which going out from solution.
2. in solution runs some type of oxidation, which decreases COD value because ratio of C:H:O increase in favor of oxygen. (E.g. values of Theoretical oxygen demand: ethanol C2H5OH has TOD 2,08 and Acetic acid CH3COOH has TOD only 1,07.)
3. in solution runs mineralization process, which transforms organic compounds into CO2 and water.
When you have on stock solution e.g. ethanol and there runs some oxidation, new formed acetic acid has lower value of COD.
However, which is second part of your original question, when you have e.g. benzo(a)pyrene and you measured value of COD, that value will be lower before oxidation, because e.g. polyaromatic compounds are badly to oxidize under COD condition. After treatment with some oxidizing agent (e.g. HO radical) you obtained molecule, which you can better oxidize under COD condition.