Rate constant K at 25°C; ultimate BOD

Check of the data

For a first-order reaction, $y_t=L(1-10^{-kt})$, the ratio $y_5/y_1$ cannot exceed 5 (it is 5 only when $k\to0$). Here

$$\frac{y_5}{y_1}=\frac{293.55}{56.15}=5.228>5$$

so the two readings cannot both belong to a single first-order curve; one of the printed values is doubtful. Therefore $K_{20}$ cannot be found from the pair. Instead, the more reliable 5-day value is used together with the standard rate constant $K_{20}=0.1$ day⁻¹ (base 10), $\theta=1.047$.

Step 1: Rate constant at 25°C

$$K_{25}=K_{20}(1.047)^{25-20}=0.1\times1.047^{5}=0.1258\ \text{day}^{-1}$$

Step 2: Ultimate first-stage BOD

$$L=\frac{y_5}{1-10^{-K_{20}\times5}}=\frac{293.55}{1-10^{-0.5}}=\frac{293.55}{0.6838}=429.3\ \text{mg/L}$$

The ultimate BOD does not depend on temperature, so the ultimate BOD at 30°C is the same, 429.3 mg/L (only the rate constant changes with temperature; $K_{30}=0.1\times1.047^{10}=0.1583$ day⁻¹, so the 5-day BOD at 30°C would be 359.9 mg/L).

Answer: $K_{25}$ = 0.1258 day⁻¹ (base 10); ultimate BOD ≈ 429.3 mg/L.