Question

Use of methanol has been proposed to rid a wastewater of nitrate by biological denitrification to N2.

(a) Write a balanced overall equation for nitrate removal with methanol, using fs=0.36 and assuming nitrate serves as the nutrient source for bacteria.

(b) If the nitrate concentration in the wastewater equals 100mg/l, what concentration of methanol must be added for complete nitrate removal?

(c) What percentage of the nitrate-nitrogen is used for cell synthesis?

please let me know the process

Answer 1

(a) Denitrification is basically biological oxidation of many organic substrates using nitrate (NO₃^-) or NO₂^- (nitrite) as e- acceptor instead of oxygen (under anoxic conditions). The nitrate reduction involve the following reduction steps from nitrate to nitrite, to nitric oxide to nitrous oxide, and in the end to nitrogen gas

NO₃^- - NO₂^- - NO - N₂O - N₂ .

The balanced overall equation for nitrate removal with methanol, using fs=0.36 and assuming nitrate serves as the nutrient source for bacteria is-

NO₃^- + 1.08CH₃OH + H⁺ = 0.47N₂ + 0.76CO₂ + 2.44H₂O + 0.065C₅H₇NO₂, where is C₅H₇NO₂ bacterial cell mass synthesized during denitrification.

(b) From above balanced equation we know for removal of 1 mol of nitrate (or 1*mol wt of nitrate, grams of nitrate) we need 1.08 mols of methanol (or 1.08*mol wt of methanol, grams of methanol).

So, 1.08*32 = 34.6 g of methanol is required to denitrify 1*62 = 62 g of NO₃^- . Therefore for 100mg/L of nitrate, we need 100*34.6/62 = 55.8 mg/L of methanol.

(c) Now from balanced equation in (a) we know that, denitrification of I mol of nitrate produces 0.47 mols of N₂, and rest goes into the synthesis of cellular mass. Or we can say 14 g of nitrate-N produces 0.47*28 = 13.16 g of N₂ gas and therefore, 14 - 13.16 = 0.84 g of nitrate-N goes in cell synthesis. In percentage it is 100*0.84/14 = 6% of nitrate-N.

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