Question

A biological reactor is getting wastewater containing 1 mole/L acetate ions as carbon source. The following reaction takes place in the bio-reactor: \[ 0.125CH_3COO^- + 0.0295NH_4^+ + 0.103O_2 \;\;\Rightarrow\;\; 0.0295C_5H_7O_2N + 0.0955H_2O + 0.095HCO_3^- + 0.007CO_2 \] Assume that all acetate ions are consumed and ammonia serves as a nutrient source. Given that 1 g acetate exerts 1.07 g COD; 1 mole bacteria = 113 g VSS; 1 mole acetate ion = 59 g. Value of observed yield is ________________________ (in g VSS/g COD, rounded off to two decimal places).

Hint:

Observed yield is calculated as biomass produced per unit COD consumed. Always use stoichiometric coefficients and molecular weights carefully.

Answer 1

Correct Answer: 0.4

Step 1: Moles of acetate consumed.
1 mole/L acetate ion is consumed = 59 g acetate.

Step 2: COD exerted.
1 g acetate = 1.07 g COD.
Therefore, 59 g acetate = \( 59 \times 1.07 = 63.13 \, \text{g COD} \).

Step 3: Biomass produced.
From stoichiometry, 0.0295 mol biomass (C\(_5\)H\(_7\)O\(_2\)N) is formed per 1 mol acetate.
Mass of biomass = \( 0.0295 \times 113 = 3.3335 \, \text{g VSS} \).

Step 4: Observed yield.
\[ Y = \frac{\text{Mass of VSS produced}}{\text{Mass of COD consumed}} = \frac{3.3335}{6.813} = 0.49 \, \text{g VSS/g COD} \] Final Answer: \[ \boxed{0.49 \, \text{g VSS/g COD}} \]