Question

The theoretical aerobic oxidation of biomass (C₅H₇O₂N) is
\[ \mathrm{C_5H_7O_2N + 5\,O_2 \rightarrow 5\,CO_2 + NH_3 + 2\,H_2O} \] 

Given a first-order biochemical oxidation with $k=0.23\ \text{d}^{-1}$ at $20^{\circ}$C (base $e$) and neglecting the second-stage oxygen demand, find the ratio BOD₅ at $20^{\circ}$C to TOC (round to two decimals). 

Atomic weights: C=12, H=1, O=16, N=14 g/mol

Hint:

For BOD problems with given stoichiometry and first-order kinetics:
1) Compute the ultimate first-stage O$_2$ demand from the reaction.
2) Convert to BOD$_t$ using $L_0(1-e^{-kt})$.
3) TOC is simply the mass of carbon in the molecule (moles of C $\times$ 12).

Answer 1

Correct Answer: 1.82

Step 1: Ultimate first-stage oxygen demand per mole of biomass
From the stoichiometry, 1 mol C₅H₇O₂N requires 5 mol O₂.
Mass of O₂ demanded (ultimate, first-stage): L₀ = 5 × 32 = 160 g O₂ per mol biomass.
 

Step 2: BOD over 5 days for first-order kinetics.
BOD_t = L₀(1-e^-kt) with k=0.23 d^-1 and t=5 d.
kt=1.15 ⇒ e^-1.15≈0.316 ⇒ 1-e^-1.15≈0.684.
Thus, BOD₅ = 160 × 0.684 ≈ 109.44 g O₂ per mol biomass.
 

Step 3: Total organic carbon (TOC) per mole of biomass.
Moles of C in C₅H₇O₂N = 5 ⇒ mass of organic carbon = 5×12 = 60 g C per mol.
 

Step 4: Ratio.
\[ \frac{\text{BOD}_5}{\text{TOC}}=\frac{109.44}{60}\approx 1.824 \ \Rightarrow\ \boxed{1.82}\ (\text{to two decimals}). \]