An industrial wastewater flowing at a rate of 37,000 m3/day is contaminated by 1

Question

An industrial wastewater flowing at a rate of 37,000 m3/day is contaminated by 187 mg/L of propanol – CH3CH2CH2OH. As an environmental consultant, you propose to treat the water in an activated sludge process in which propanol is oxidized by oxygen into carbon dioxide – CO2. The reaction is highly favorable and assumed to be complete. a) Equilibrate the half-reactions and overall reaction that are expected to occur during the treatment process. b)Determine the theoretical quantity of oxygen (in mol/day) required for the complete oxidation of propanol.

Explanation / Answer

Answer – We are given, wastewater flowing at a rate = 37,000 m3/day

[propanol] = 187 mg/L

a)The equilibrate the half-reactions

Oxidation –

2 CH3CH2CH2OH + 10 H2O -----> 6 CO2 + 36H+ + 36e-

Reduction -

9 O2 + 36H+ + 36e- ----> 18 H2O

Overall reaction –

2 CH3CH2CH2OH + 9 O2 ------> 6 CO2 + 8 H2O

b) For 1 day wastewater flowing = 37000 m3

we need to convert the m3 to dm3

we know

1 m3 = 1000 dm3

37000 m3 = ?

= 3.70*107 dm3

= 3.70*107 L

For 1 L = 187 mg of CH3CH2CH2OH

So, 3.70*107 L = ?

= 6.92*109 mg of CH3CH2CH2OH

= 6.92*106 g of CH3CH2CH2OH

Moles of CH3CH2CH2OH = 6.92*106 g / 60.096 g.mol-1

                                           = 1.15*105 moles

Form the overall balanced reaction –

2 moles of CH3CH2CH2OH = 9 moles of O2

So, 1.15*105 moles of CH3CH2CH2OH = ?

= 5.18*105 moles of O2

So, the theoretical quantity of oxygen (in mol/day) required for the complete oxidation of propanol is 5.18*105 moles/day of O2

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