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

Question 6 of 10: Henry's Law Constant for Benzene and Its Use in Remediation Screening

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 6: Henry's Law Constant for Benzene and Its Use in Remediation Screening (10 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. Benzene, $T = 25\,{}^{\circ}\text{C} = 298.15\ \text{K}$; open-book literature value for benzene's Henry's law constant, $H \approx 5.55\times10^{-3}\ \text{atm}\cdot\text{m}^3/\text{mol}$ at 25 °C.

Find. (a) $H$ in atm·m³/mol; (b) the dimensionless form $H'$; (c) what this value implies for remediation-method selection.

Approach. Look up benzene's Henry's constant (open-book data), then convert to dimensionless form via $H' = H/(RT)$, and compare $H'$ against the standard air-stripping screening rule of thumb.

  1. (a) Henry's constant, conventional units. From standard tables, benzene at 25 °C has $$\boxed{H \approx 5.55\times10^{-3}\ \text{atm}\cdot\text{m}^3/\text{mol}}$$
  2. (b) Convert to dimensionless form. Using $H' = H/(RT)$ with $R = 8.206\times10^{-5}\ \text{atm}\cdot\text{m}^3/(\text{mol}\cdot\text{K})$, $$H' = \frac{5.55\times10^{-3}}{(8.206\times10^{-5})(298.15)} = \frac{5.55\times10^{-3}}{0.02447} = \boxed{0.227}$$

c. The rule of thumb used for screening air-stripping feasibility (Kavanaugh & Trussell) is that a dimensionless $H'$ above about 0.01 (equivalently $H$ above roughly $1\times10^{-3}\ \text{atm}\cdot\text{m}^3/\text{mol}$) marks a compound as volatile enough for air stripping to be an efficient, low-cost mass-transfer process. Benzene's $H' \approx 0.227$ is more than twenty times that threshold, so benzene partitions strongly into the gas phase and is an excellent candidate for physical volatilization technologies — air stripping (packed-tower or diffused-air) or soil-vapour extraction — rather than adsorption-based methods such as granular activated carbon, which are the preferred choice for low-volatility, high-Kow compounds instead. This is exactly the technology basis for the air-stripping tower design in Question 8, which treats this same contaminant.

Question 6 — final results
QuantityValue
Henry's constant, $H$5.55×10−3 atm·m³/mol
Dimensionless Henry's constant, $H'$0.227
Remediation implication$H'\gg 0.01$ → air stripping/SVE favoured over carbon adsorption