Biochemical oxygen demand is a measure of the potential damage that can be done to the dissolved oxygen content of a body of water by organic waste. A BOD is the amount of O2 required to degrade or decompose the wastes. Its name comes from the "demand" that organic carbon makes for O2 through biochemical (enzyme) oxidation.
BOD - demanded oxygen - can be likened to a debt that must be paid. Dissolved oxygen (DO) can be thought of as the currency used to pay the debt. If DO (assets) exceeds BOD (debt), aerobic conditions (economic solvency) will prevail. If BOD exceeds DO, bankruptcy may ensue. Bankruptcy, in this case, is the depletion of water's chief asset, O2. Over a period of time, the water's oxygen level is usually replenished by oxygen from the air. The length of this time period depends on the severity of the initial loss of oxygen.
If we assume that the part of the organic waste that is being degraded is carbon, C ( a good assumption), then we can write the a chemical formula.
If we were to do a mass analysis (i.e. gram for gram, pound for pound, etc.) of this equation, we would find that for every 3 parts of C, we would need 8 parts of O2. If you're thinking that it looks like it should be a one for one deal, you're absolutely correct. It does in fact require one MOLECULE of carbon to react with one MOLECULE of oxygen, but if we would weigh those molecules we'd see that based on mass, we'd get the 3:8 ratio. Using this information answer the following questions in the Multi-part Assignment.
Part 1: Consider a small lake, 0.5 mile across and 8 ft deep, that contains a total of 32 tons of dissolved oxygen. Calculate the BOD (i.e. calculate the amount of oxygen required to degrade the amount of organic waste given) if I dump 6 tons of carbon-based waste into the lake.
Part 2: Repeat the same calculation, but this time assume we dump 60 tons of carbon-based waste into the lake.
3: Discuss the differences between the two cases in terms of how much oxygen each case requires versus how much oxygen is available. Discuss any possible effects on the lake and its inhabitants.
The overall equation did not show up, however I am going to guess that you are talking about there needing one mole of oxygen gas, O2,(molar mass = 32 g/mol) for every mole of carbon, C, (molar mass = 12). When you divide both molar masses buy the greatest common factor of 4, you get your 8:3 ratio. This means that for every 3 grams (or pounds) ...
This solution offers help with environmentaly related chemical equations.