---
title: "Ultimate BOD and rate constant"
subject: "Sanitary Engineering"
chapter: 5
key: "semester_6/sanitary_engineering:bod-ultimate-rate-constant-100-210"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-ultimate-rate-constant-100-210
site: BCT Engineering
---

# Ultimate BOD and rate constant

Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater

- Asked in: 2076 Baisakh, 8 marks
- Video: [Special Lecture on BOD and Ultimate BOD.](https://www.youtube.com/watch?v=ENZCCOfjzZE) — Easy Learning
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-ultimate-rate-constant-100-210
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-5

## Question

If a water sample has BOD1 at 20°C = 100 mg/l and BOD5 at 25°C = 210 mg/l; what will be the ultimate BOD and rate constant? Assume suitable data if necessary. Determine BOD4 at 22°C of the same water sample.

## Short answer

Solving the system y1=100=L(1-10^(-k20)), y5=210=L(1-10^(-5·k25)) with k25=1.258k20 gives k20=0.273/day and ultimate BOD L=214.1 mg/L. Then k22=0.273×1.047^2=0.2997/day, and BOD4 at22°C =214.1(1-10^(-4·0.2997)) =200.5 mg/L.

## Answer

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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is k20 found when y1 at 20°C and y5 at 25°C are given?
   A. Divide the two BOD equations to remove L, with k25 = 1.2582 k20, and solve by trial
   B. Add the two equations and solve directly
   C. Assume k20 = 0.1 and ignore the data
   D. Use only y5 and set L = 210
   Answer: A. The ratio of the two equations contains only k20, so it is solved by trial. (Key point: Method for unknown k)

2. What is k25 in terms of k20?
   A. k25 = k20 x 1.047^25
   B. k25 = k20 x 1.047^5 = 1.2582 k20
   C. k25 = k20 / 1.2582
   D. k25 = k20 x 1.047^20
   Answer: B. The exponent is T - 20 = 5. (Key point: Temperature factor)

3. What is the ultimate BOD found?
   A. About 100 mg/L
   B. About 210 mg/L
   C. About 214 mg/L
   D. About 428 mg/L
   Answer: C. L = 100 / (1 - 10^(-0.273)) = 214.1 mg/L. (Key point: Ultimate BOD)

4. How is BOD4 at 22°C found?
   A. Use k20 and a new L for 22°C
   B. Use k25 with y4 = L x 10^(-4k)
   C. Multiply y5 at 25°C by 4/5
   D. Use k22 = k20 x 1.047^2 with the same L in y4 = L(1 - 10^(-4 k22))
   Answer: D. L is independent of temperature; only k is corrected, giving y4 about 200.5 mg/L. (Key point: Final BOD4 at 22°C)

## Related in Sanitary Engineering

- [Rate constant K at 25°C; ultimate BOD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-k-ultimate-293-56) — Sanitary Engineering, chapter 5
- [BOD1 at 22°C to BOD5 at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-1day-22c-310-to-5day-30c) — Sanitary Engineering, chapter 5
- [BOD5 and ultimate BOD, 5% dilution](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-dilution-5pct-blank) — Sanitary Engineering, chapter 5
- [BOD, COD; ultimate BOD of 480 mg/l](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-cod-ultimate-bod-480) — Sanitary Engineering, chapter 5
- [Rate constant and remaining BOD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-rate-250-325) — Sanitary Engineering, chapter 5
- [BOD1 at 23°C to BOD5 at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-1day-23c-105-to-5day-30c) — Sanitary Engineering, chapter 5

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