Question

Determine the Theorethical Oxygen Demand(ThOD) for sewage containing 150 mg/L of CH₂(NH₂)COOH and 40 mg/L of NH₃.

Find,

(1) mg/L of oxygen for both carbon and nitrogen compounds

(2) grams of oxygen per gram of CH2(NH2)COOH (oxygen used only for carbon oxidation)

(3) grams of oxygen per gram of NH3 (oxygen used only for nitrification)

Answer 1

GLYCINE OXIDATION SCHEME

NH2CH2COOH + ½ o2→OHC-COOH (GLYOXYLIC ACID)+NH3

OHC-COOH +[O](1/2 o2) →HCOOH(FORMATE)+[O] (1/2 O2)→CO2+H2O (COMPLETE OXIDATION)

Total o2 required for carbon oxidation=3/2=1.5 mole O2

NITRIFICATION EQUATION

1)NH3+1.5O2→NO2- +H+ +H2O

2)NO2- +1/2O2→NO3-

TOTAL O2 REQUIRED PER MOLE NH3 =2 MOLE O2

GIVEN

GLYCINE CONCENTRATION= 150 mg/L of CH2(NH2)COOH

Molar conc=(0.150 g/molar mass) per L=0.150 g/75.07 g/mol=0.002 mol/L

NH3 CONC=40 mg/L

Molar conc=0.040g/17.031g/mol=0.00235 mol/L

(1) mg/L of oxygen for both carbon and nitrogen compounds

Total moles of O2 for 1 mole glycine oxidation=1.5 mole(C-oxidation of glycine)+2 mole O2 (Nh3 oxidation obtained from glycine) =3.5 mol

For 0.002mol of glycine ,moles of O2 required=0.002 *3.5 mol=0.007 moles of O2=0.007 moles*32g/mol=0.224 g=224 mg

So for 0.002 mol/L of glycine mass of O2 required=224g/L

For 1 mole NH3 oxidation)=2 mole O2 required

So for 0.00235 mol NH3 oxidation=0.00235 *2=0.0047 moles

0.00235 mol/L NH3 oxidation requires =0.0047 moles/L O2=(0.0047 moles*32g/mol)/L=0.150 g/L=150 mg/L

Total O2 required=224 g/L+150g/L=374 g/L

(2) grams of oxygen per gram of CH2(NH2)COOH (oxygen used only for carbon oxidation)

Solved in part (1),

Total moles of O2 for 1 mole glycine oxidation=1.5 mole(C-oxidation of glycine)

0.002 mol/L of glycine requires 1.5*0.002=0.003 mol/L of O2=(0.003 mol*32g/mol)/L=0.096 g/L=96 mg/L

0.002 moles*75.07 g/mol=0.150 g glycine requires 0.096 g O2

(3) grams of oxygen per gram of NH3 (oxygen used only for nitrification)

For 1 mole NH3 oxidation)=2 mole O2 required

So for 0.00235 mol NH3 oxidation=0.00235 *2=0.0047 moles

0.00235 mol/L NH3 oxidation requires =0.0047 moles/L O2=(0.0047 moles*32g/mol)/L=0.150 g/L=150 mg/L

Consequently,0.00235 mol*17.031 g/mol=0.04 g of NH3 requires 0.150 g O2