---
title: "Rate constant and remaining BOD"
subject: "Sanitary Engineering"
chapter: 5
key: "semester_6/sanitary_engineering:bod-rate-250-325"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-rate-250-325
site: BCT Engineering
---

# Rate constant and remaining BOD

Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater

- Asked in: 2076 Bhadra, 4 marks
- Video: [BOD equation derivation|BOD Rate constant|Deoxygenation constant|coronaclasses|civil](https://www.youtube.com/watch?v=m8jV_mctuL8) — ISHA MISHRA
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-rate-250-325
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-5

## Question

A wastewater sample was incubated at 25°C. 5 day BOD of sample was equal to 250 mg/l and 11 day BOD of sample was found to be 325 mg/l. Calculate the rate reaction constant and organic matter remaining in sample after 6 days at 30°C.

## Short answer

From y5/y11 ratio, solving (1-10^(-11k))/(1-10^(-5k))=1.3 yields k25=0.1118/day (base10). Ultimate BOD L=345.4 mg/L. At30°C, k30=0.1406/day; remaining organic matter after6 days = L·10^(-k30·6) =49.5 mg/L.

## Answer

Given at 25°C: $y_5=250$ mg/L, $y_{11}=325$ mg/L. Assume a first-order reaction in base 10, $y_t=L(1-10^{-kt})$, and $\theta=1.047$.

#### Step 1: Rate constant at 25°C
Divide the two equations to remove $L$:

$$
\frac{y_{11}}{y_5}=\frac{1-10^{-11k}}{1-10^{-5k}}=\frac{325}{250}=1.3
$$

Solving by trial: $k_{25}=0.1$ gives 1.346, $k_{25}=0.2$ gives 1.104; refining gives

$$
k_{25}=0.1118\ \text{day}^{-1}\ (\text{base }10)
$$

#### Step 2: Ultimate BOD
$$
L=\frac{250}{1-10^{-0.1118\times5}}=345.4\ \text{mg/L}
$$

#### Step 3: At 30°C
$$
k_{30}=k_{25}(1.047)^{5}=0.1118\times1.2582=0.1406\ \text{day}^{-1}
$$

Organic matter remaining after 6 days (the ultimate BOD is the same at any temperature):

$$
L_6=L\cdot10^{-k_{30}\times6}=345.4\times10^{-0.8439}=49.5\ \text{mg/L}
$$

**Answer: $k_{25}$ = 0.1118 day⁻¹; organic matter remaining after 6 days at 30°C = 49.5 mg/L** (BOD exerted = 295.9 mg/L).

## Check yourself

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

1. How is the rate constant found from y5 = 250 and y11 = 325?
   A. Divide y11 by y5 to remove L: (1 - 10^(-11k))/(1 - 10^(-5k)) = 1.3 and solve by trial
   B. Subtract the two values and divide by 6
   C. Use k = ln(325/250)/6
   D. Set L = 325 and solve for k
   Answer: A. The ratio of the two equations has only k unknown. (Key point: Eliminating L)

2. Which is the ultimate BOD?
   A. About 250 mg/L
   B. About 345 mg/L
   C. About 325 mg/L
   D. About 575 mg/L
   Answer: B. L = 250 / (1 - 10^(-0.559)) = 345.4 mg/L. (Key point: Ultimate BOD)

3. How is k30 obtained from k25?
   A. k30 = k25 x 1.047^30
   B. k30 = k25 / 1.047^5
   C. k30 = k25 x 1.047^5
   D. k30 = k25 x 1.047^10
   Answer: C. Temperature difference is 30 - 25 = 5, giving 0.1406 per day. (Key point: Temperature correction)

4. Which formula gives the organic matter remaining after 6 days?
   A. L6 = L(1 - 10^(-k30 x 6))
   B. L6 = y11 - y5
   C. L6 = L x 10^(-k25 x 6)
   D. L6 = L x 10^(-k30 x 6)
   Answer: D. The remaining matter is L10^(-kt) = 49.5 mg/L; the exerted part is L(1 - 10^(-kt)). (Key point: Remaining organic matter)

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