Contamination Sources, Hydrolysis Decay, Henry's Law Partitioning, and a Gasoline Mass Balance: 18-Geol-B1 Contaminant Hydrogeology, May 2017, Q2

Question 2: Contamination Sources, Hydrolysis Decay, Henry's Law Partitioning, and a Gasoline Mass Balance (equal value)

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.

Part (a) — six source types of groundwater contamination. (1) Leaking underground storage tanks and pipelines (petroleum, solvents). (2) Septic systems and on-site sewage disposal. (3) Landfills and waste disposal sites (municipal and industrial). (4) Agricultural activities — fertilizer and pesticide application, animal waste lagoons/manure storage. (5) Surface impoundments, lagoons, and industrial spills/leaks at manufacturing or chemical-storage facilities. (6) Abandoned or improperly constructed/sealed wells and boreholes that provide a direct conduit between the surface (or a contaminated shallow zone) and deeper aquifers; road salt/de-icing chemical application is an additional common source in Canadian practice.

Given. (b) First-order hydrolysis rate coefficient $k=0.001\ \text{day}^{-1}$. (c) Dissolved benzene $C_w=260\ \mu\text{g/L}$, Henry's constant $H=5.5\times10^{-3}\ \text{atm-m}^3/\text{mol}$, molecular weight $MW=78.1\ \text{g/mol}$, pure-liquid benzene vapour pressure $P_v^{o}=0.1\ \text{atm}$. (d) Release volume 1000 gal gasoline, gasoline density $\rho=0.9\ \text{g/mL}$, benzene mass fraction $\approx1\%$, plume $100\times50\times10\ \text{ft}$, average dissolved concentration $C=0.1\ \text{mg/L}$, porosity $n=0.3$; no volatilization, biodegradation, or sorption loss.

Find. (b) Time for 99% hydrolysis. (c) Equilibrium soil-gas partial pressure from the dissolved-phase concentration, and the partial pressure if benzene were present as pure NAPL. (d) Mass of benzene released, and how that mass splits between the dissolved and NAPL phases.

Approach. (b) Invert the first-order decay law for the elapsed time at 1% remaining. (c) Apply Henry's law to the dissolved concentration for the equilibrium vapour partial pressure, then apply Raoult's law (mole fraction ≈ 1 for a nearly-pure liquid) for the pure-NAPL case. (d) Convert the total release to a benzene mass, compute the mass held in the plume's pore water from its volume/porosity/concentration, and take the remainder (by the no-loss assumption) as NAPL mass.

Part (b) — time to 99% hydrolysis. First-order decay: $C/C_0=e^{-kt}$; 99% hydrolyzed means 1% remains, so $C/C_0=0.01$. $$t=\frac{-\ln(0.01)}{k}=\frac{\ln(100)}{0.001}=\boxed{4605\ \text{days}\ (12.6\ \text{years})}.$$
Part (c) — equilibrium soil-gas partial pressure from dissolved phase. Convert $C_w$ to molar concentration: $C_w=260\ \mu\text{g/L}=0.260\ \text{g/m}^3$, so $C_w=0.260/78.1=3.329\times10^{-3}\ \text{mol/m}^3$. Henry's law: $$P_g=H\,C_w=(5.5\times10^{-3})(3.329\times10^{-3})=\boxed{1.83\times10^{-5}\ \text{atm}}.$$
Part (c), continued — partial pressure from pure-liquid NAPL. When benzene is present as a separate (pure or near-pure) liquid phase, Raoult's law gives $P=x_{benzene}P_v^{o}$ with $x_{benzene}\approx1$, so $$P=\boxed{0.1\ \text{atm}},$$ roughly 5,500× higher than the dissolved-phase equilibrium pressure — a direct illustration of why even a small residual NAPL "hot spot" dominates the vapour-phase risk pathway compared with a dissolved plume of the same footprint.
Part (d) — total benzene released. $1000\ \text{gal}\times3.7854\ \text{L/gal}=3785.4\ \text{L}$ of gasoline; mass $=3785.4\ \text{L}\times1000\ \text{mL/L}\times0.9\ \text{g/mL}=3.407\times10^{6}\ \text{g}$. At 1% benzene by mass: $$m_{benzene}=0.01\times3.407\times10^{6}=\boxed{34{,}069\ \text{g}\ (34.1\ \text{kg})}.$$
Dissolved-phase mass. Plume volume $=100\times50\times10\ \text{ft}^3=(30.48)(15.24)(3.048)\ \text{m}^3=1415.8\ \text{m}^3$. Pore-water volume $=nV=0.3\times1415.8=424.8\ \text{m}^3=424{,}800\ \text{L}$. $$m_{dissolved}=C\,V_w=(0.1\ \text{mg/L})(424{,}800\ \text{L})=42{,}480\ \text{mg}=\boxed{42.5\ \text{g}}.$$
NAPL-phase mass (by conservation). With no volatilization, biodegradation, or sorption loss, every gram of benzene is either dissolved or still present as NAPL: $$m_{NAPL}=m_{benzene}-m_{dissolved}=34{,}069-42.5=\boxed{34{,}026\ \text{g}\ (34.0\ \text{kg})}.$$ Only about 0.12% of the released benzene has dissolved after one year — benzene's aqueous solubility (~1,780 mg/L pure-phase) limits how fast NAPL can partition into passing groundwater, so the great majority of the mass remains as residual/free-phase NAPL long after release, which is exactly why NAPL recovery (not just dissolved-plume monitoring) is central to remediating a site like this.

Question 2 — Final Results
Item	Result
2(b) Time to 99% hydrolysis	4605 days (12.6 yr)
2(c) Equilibrium vapour pressure, dissolved-phase source	$1.83\times10^{-5}$ atm
2(c) Vapour pressure, pure-NAPL source	0.1 atm
2(d) Total benzene released	34,069 g (34.1 kg)
2(d) Dissolved-phase mass	42.5 g (0.12% of release)
2(d) NAPL-phase mass	34,026 g (99.88% of release)

Concept & Theory

Topic. Contamination source inventory, first-order hydrolysis, Henry's law/Raoult's law vapour partitioning, and a conservative (no-loss) NAPL/dissolved-phase mass balance.

Key relations. $C/C_0=e^{-kt}$; $P_g=H\,C_w$ (dilute aqueous source, Henry's law); $P=x_i P_i^{o}$ (Raoult's law, near-pure liquid source); mass balance $m_{released}=m_{dissolved}+m_{NAPL}$ (no sorption/volatilization/biodegradation losses).

Why this works. Henry's law is the dilute-solution limit of vapour-liquid equilibrium and applies once benzene is present only as a trace dissolved species; once a separate benzene-rich liquid phase (NAPL) exists, the relevant equilibrium is instead governed by Raoult's law referenced to the pure-component vapour pressure, which is why the same compound produces two very different partial pressures depending on which phase is source. The mass-balance step works because benzene's aqueous solubility limit means only a small fraction of an NAPL release can dissolve into a fixed pore-water volume at a time; the remainder simply persists as NAPL until it dissolves, volatilizes, biodegrades, or is recovered.

Common pitfall. Forgetting to convert $\mu\text{g/L}$ to mol/m³ (or mixing L and m³ units) before applying Henry's law — a factor-of-1000 slip here silently propagates into the boxed pressure. A second pitfall is treating "no sorption/volatilization/biodegradation" as meaning the dissolved concentration equals the total average concentration in the whole plume volume, rather than in the pore-WATER volume only ($V_w=nV$, not $V$).

Source. Fetter, Contaminant Hydrogeology Ch.3 (Henry's law, Raoult's law, NAPL fate) & Ch.1 (contamination sources); course formula sheet (first-order hydrolysis).

Question 2 of 5: Contamination Sources, Hydrolysis Decay, Henry's Law Partitioning, and a Gasoline Mass Balance

Question 2: Contamination Sources, Hydrolysis Decay, Henry's Law Partitioning, and a Gasoline Mass Balance (equal value)