Question

The theoretical aerobic oxidation of biomass (C$_5$H$_7$O$_2$N) 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 BOD$_5$ 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

Step 1: Ultimate first-stage oxygen demand per mole of biomass.
From the stoichiometry, $1$ mol C$_5$H$_7$O$_2$N requires $5$ mol O$_2$.
Mass of O$_2$ demanded (ultimate, first-stage): $L_0 = 5 \times 32 = 160$ g O$_2$ per mol biomass.

Step 2: BOD over 5 days for first-order kinetics.
BOD$_t = L_0\left(1-e^{-kt}\right)$ with $k=0.23\ \text{d}^{-1}$ and $t=5$ d.
$kt=1.15 \Rightarrow e^{-1.15}\approx 0.316 \Rightarrow 1-e^{-1.15}\approx 0.684$.
Thus, $\text{BOD}_5 = 160 \times 0.684 \approx 109.44$ g O$_2$ per mol biomass.

Step 3: Total organic carbon (TOC) per mole of biomass.
Moles of C in C$_5$H$_7$O$_2$N $=5$ $\Rightarrow$ mass of organic carbon $=5\times 12=60$ g C per mol.

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