02-19 & 02-22 - OXYGEN DEMAND CONCEPTS Now that we...

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OXYGEN DEMAND CONCEPTS Now that we know about the kinetics of substrate (and, therefore, Oxygen) utilization we need to think in terms of amounts. I.E. given some general organic compound, how much O 2 is needed to oxidize the carbon to CO 2 and the nitrogen to NO 3 - (CO 2 and NO 3 - being the most oxidized states of C and N)? We can think of the process of oxidation occurring in 2 stages: I . Carbonaceous Oxidation C a H b O c N d + ____________________ O 2 ___CO 2 + ________ H 2 O +___NH 3 [Note the oxidation number of N is unchanged by this reaction] 1/2[2a + 1/2(b - 3d) -c] a 1/2(b-3d) d - III
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II . Nitrogenous Oxidation ____ NH 3 + ____ O 2 ___ NO 3 - + ____ H 2 O + ____ H + Example: oxidation of acetic acid ( CH 3 COOH). a=_____ b=_____ c=_____ d=_____ CH 3 COOH + ____ O 2 _____CO 2 + _____ H 2 O 1 mole = 60g 32 g/mole d 2d d d d -III +V 2 4 2 0 2 2 2 2 moles of O 2 per mole of waste 1/2[ 2a + 1/2 (b - 3d) -c] .5{4 + .5[4-3(0)] -2} 1/2(b-3d) .5[4-3(0)]
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Example (continued) calculate 64g O 2 required to oxidize 60g CH 3 COOH = 1.067g O 2 demand g acetic acid So, if we had 100 mg/L CH 3 COOH, it's O 2 demand would be ________________mg/L. If we do a lab analysis to determine the composition of an organic compound (i.e. figure out what the coefficients a, b, c & d in C a H b O c N d are), we can figure out how much O 2 bacteria would use when they oxidize it. Right? ______________________________ 2 moles O 2 @ 32g/mole 106.7 We can convert a waste conc. (in moles/L or mg/L) to an equivalent amount of O 2 needed to oxidize it (mg O 2 demand/L).
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02-19 & 02-22 - OXYGEN DEMAND CONCEPTS Now that we...

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