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.

Answer 1

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