Ultimate BOD and rate constant

Given: $y_1$ at 20°C = 100 mg/L and $y_5$ at 25°C = 210 mg/L. Assumptions: first-order (base 10) reaction, ultimate BOD $L$ independent of temperature, and $k_T=k_{20}(1.047)^{T-20}$.

Step 1: Equations

$$\begin{aligned} 100&=L\left(1-10^{-k_{20}\times1}\right)\\ 210&=L\left(1-10^{-k_{25}\times5}\right),\qquad k_{25}=k_{20}(1.047)^{5}=1.2582\,k_{20} \end{aligned}$$

Dividing the two equations gives one unknown, $k_{20}$:

$$\frac{1-10^{-1.2582\,k_{20}\times5}}{1-10^{-k_{20}}}=2.1$$

Solving by trial (trial values of $k_{20}$ are tried until the ratio is 2.1; e.g. $k_{20}=0.2$ gives 2.560, $k_{20}=0.3$ gives 1.979):

$$k_{20}=0.273\ \text{day}^{-1},\qquad k_{25}=0.344\ \text{day}^{-1}$$

Step 2: Ultimate BOD

$$L=\frac{100}{1-10^{-0.273}}=214.1\ \text{mg/L}$$

(Check at 25°C: $y_5=214.1(1-10^{-0.344\times5})=210.0$ mg/L, as given.)

Step 3: BOD$_4$ at 22°C

$$\begin{aligned} k_{22}&=0.273\times1.047^{2}=0.2997\ \text{day}^{-1}\\ y_4&=L\left(1-10^{-k_{22}\times4}\right)=214.1\left(1-10^{-1.1986}\right)=200.5\ \text{mg/L} \end{aligned}$$

Answer: Ultimate BOD = 214.1 mg/L; rate constant $k_{20}$ = 0.273 day⁻¹ (base 10) ($k_{25}$ = 0.344); BOD₄ at 22°C = 200.5 mg/L.