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18-Env-B5 Industrial & Hazardous Waste Management · Undated paper

Question 5 of 10: First-Order Half-Life of Toluene in a Treatment Pond

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2019 — 18-Env-B5: Industrial & Hazardous Waste Management (3 hours, open book). Marks are indicated beside each question for a total of 100 marks; all ten questions are answered in full below as a complete study resource.

Reference texts: LaGrega, Buckingham & Evans, Hazardous Waste Management (2nd ed.); Nemerow & Dasgupta, Industrial and Hazardous Waste Treatment (2nd ed.); Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.); Davis & Cornwell, Introduction to Environmental Engineering (6th ed.).

Question 5: First-Order Half-Life of Toluene in a Treatment Pond (5 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. First-order decay, $C_0 = 30\ \text{mg/L}$, rate constant $k_T = 0.067\ \text{hr}^{-1}$.

Find. The half-life $t_{1/2}$ (time for $C$ to fall to $C_0/2$).

Approach. Integrate the first-order rate law and set $C/C_0 = 1/2$ to solve for the characteristic half-life; the initial concentration itself cancels out of a half-life calculation.

  1. Integrate the rate law. $dC/dt = -k_T C \Rightarrow C(t) = C_0 e^{-k_T t}$.
  2. Solve for the half-life. Setting $C(t_{1/2}) = C_0/2$, $$\frac{1}{2} = e^{-k_T t_{1/2}} \quad\Rightarrow\quad t_{1/2} = \frac{\ln 2}{k_T} = \frac{0.6931}{0.067\ \text{hr}^{-1}} = \boxed{10.3\ \text{hours}}$$

The initial concentration of 30 mg/L never enters the calculation — a first-order half-life is independent of starting concentration, which is precisely why it is the standard way to characterize this kind of removal constant. At $k_T = 0.067\ \text{hr}^{-1}$, toluene concentration would fall to 25 % ($C_0/4$) after two half-lives, roughly 20.7 hours, and to below 5 % after about 44.7 hours (four to five half-lives), giving a practical sense of the detention time a treatment pond would need to rely on this removal mechanism alone.

This half-life is a useful sizing check for a treatment pond or lagoon: if the pond's actual hydraulic residence time is much shorter than 10.3 hours, first-order biodegradation alone will not carry the bulk of the removal duty, and volatilization or adsorption to sediment would need to be relied on instead (or the pond enlarged/staged to lengthen contact time). Conversely, a pond already sized for a day or more of residence time has ample margin — several half-lives — to drive toluene well below typical discharge limits through this pathway alone, provided the rate constant used here genuinely reflects the pond's temperature and biomass conditions rather than a generic literature value.

Question 5 — final results
QuantityValue
Half-life, $t_{1/2}$10.3 hours