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Environmental Chemistry Lecture Notes (Bullet Points)

Problem Set Review: Mass Balances, PFR, and Environmental Chemistry

  • The lecture covers homework solutions for mass-balance problems, then introduces environmental chemistry with many math-heavy concepts.
  • Key themes: steady-state vs non-steady-state, non-conservative pollutants, CSTR and PFR models, unit conversions, chemical equilibria, acid-base chemistry in rainwater and oceans, and practical environmental implications.

CSTR Problem 1: Steady-state with a non-conservative pollutant ( lagoon )

  • Given:
    • Lagoon volume: V = 1200\ \text{m}^3
    • Inflow rate: Q_{\text{in}} = 100\ \text{m}^3\,\text{d}^{-1}
    • Outflow rate: Q{\text{out}} = Q{\text{in}} = 100\ \text{m}^3\,\text{d}^{-1}
    • Incoming concentration: C_{\text{in}} = 10\ \text{mg L}^{-1}
    • Decay rate (non-conservative): k = 0.2\ \text{d}^{-1}
  • Objective:
    • (A) Find outlet concentration C_{\text{out}} at steady state.
    • (B) If input concentration jumps to C_{\text{in}} = 100\ \text{mg L}^{-1}, determine effluent concentration after 7 days (non-steady state).
  • Model at steady state with a first-order decay (non-conservative) for a CSTR:
    • Mass balance yields
      C{\text{out}} = \frac{Q{\text{in}} C{\text{in}}}{Q{\text{out}} + kV}.
  • Part (A) calculation:
    • Denominator: Q_{\text{out}} + kV = 100 + (0.2)(1200) = 100 + 240 = 340\
    • Numerator: Q{\text{in}} C{\text{in}} = 100 \times 10 = 1000
    • C_{\text{out}} = \frac{1000}{340} = 2.94\ \text{mg L}^{-1}
  • Answer (A): C_{\text{out}} = 2.94\ \text{mg L}^{-1}
  • Part (B) setup:
    • After the change, the new steady-state concentration is
      C{\infty} = \frac{Q{\text{in}} C{\text{in,new}}}{Q{\text{out}} + kV} = \frac{100\times 100}{340} = 29.41\ \text{mg L}^{-1}.
    • The initial lagoon concentration is the solution from Part (A): C_0 = 2.94\ \text{mg L}^{-1}.
    • For non-steady-state, the transient solution for a CSTR with first-order decay is
      C(t) = C{\infty} + (C0 - C{\infty}) \exp\left[-\left(\frac{Q{\text{out}}}{V} + k\right) t\right].
    • Plug in numbers: \frac{Q_{\text{out}}}{V} = \frac{100}{1200} = 0.0833\ \text{d}^{-1},\quad k = 0.2\ \text{d}^{-1}
    • Sum: 0.0833 + 0.2 = 0.2833\ \text{d}^{-1}
    • Time: t = 7\ \text{days}
    • Exponential term: \exp(-0.2833 \times 7) \approx \exp(-1.983) \approx 0.1368
    • Then
      C(7) = 29.41 + (2.94 - 29.41) \times 0.1368 \approx 29.41 - 3.63 \approx 25.8\ \text{mg L}^{-1}.
  • Answer (B): Approximately C(7\ \text{days}) \approx 25.8\ \text{mg L}^{-1}.
  • Notes:
    • If you’re stuck, double-check units when converting time constants (d^-1) and volumes (m^3).
    • The key is using the steady-state expression for C_out and then the transient solution for a step-change input.

CSTR Problem 2: Plug Flow Reactor with ozone decay

  • Setup:
    • Ozone injected into a pipe (PFR) for disinfection.
    • Pipe diameter: d = 3\ \text{ft}; length: L = 3400\ \text{ft}
    • Flow rate: Q = 10{,}000\ \text{gpm}
    • Desired exit concentration: C{\text{out}} = 1\ \text{mg L}^{-1}; find C{\text{in}} = C_0.
  • Given that ozone decays with first-order kinetics: the problem provides that concentration decreases by 50% in 12 minutes:
    • For a PFR, the concentration at time t is
      C(t) = C_0 e^{-kt}.
    • Solve for k using 50% decay in 12 min:
      0.5 = e^{-k(12\ \text{min})} \Rightarrow k = -\frac{1}{12}\ln(0.5) \approx 0.0578\ \text{min}^{-1}.
  • Time for water to traverse the pipe (residence time):
    • Volume of the pipe: V_{\text{pipe}} = A L = \left(\frac{\pi d^2}{4}\right) L.
    • With d = 3 ft, A = (π/4) d^2 ≈ 7.0686 ft^2, so V ≈ 7.0686 × 3400 ≈ 24{,}033\ \text{ft}^3.
    • Flow in ft^3/min: 1 ft^3 = 7.48 gal, so Q = 10{,}000 / 7.48 ≈ 1337.8 ft^3/min.
    • Residence time: t = \frac{V}{Q} \approx \frac{24{,}033}{1337.8} \approx 18\ \text{min}.
  • Required Cin from Cout:
    • For a PFR: C{\text{out}} = C0 e^{-kt} \Rightarrow C0 = C{\text{out}} e^{kt}.
    • Compute kt: k t = (0.0578\ \text{min}^{-1})(18\ \text{min}) \approx 1.04.
    • Therefore, C_{\text{in}} = 1 \times e^{1.04} \approx 2.83\ \text{mg L}^{-1}.
  • Answer: The concentration at the pipe inlet should be approximately C_{\text{in}} \approx 2.83\ \text{mg L}^{-1}.
  • Notes:
    • Important steps: compute k from half-life, compute residence time t from volume and Q, then use the PFR decay equation to back-calculate Cin.
    • A unit care point: convert all volumes and flow rates to consistent units before calculating t.

Quick refresher: key units, conversions, and constants

  • Concentrations and units
    • For chemical reactions, use molar concentrations: [A] = \text{mol L}^{-1}. At standard conditions, there are conversions between molarity and other units.
    • Water density ~ 1 kg L^{-1} implies 1 L of water ≈ 1000 g, so
      1\ \text{mg L}^{-1} \approx 1\ \text{ppm} for water-like substances.
    • For gases, common units are ppm or ppb; 1 ppm (gas) = 1 mol gas per 10^6 mol air.
  • Ideal gas law (environmental context)
    • General form: pV = nRT,\quad P\text{ in atm},\quad V\text{ in L},\quad n\text{ in mol},\quad R = 0.082057\ \text{L atm mol}^{-1}\text{K}^{-1}.
    • Volume per mole at STP (0 °C, 1 atm): V_{\text{STP}} = 22.4\ \text{L} = 0.0224\ \text{m}^3\text{ per mole}.
    • At 25 °C (298.15 K) and 1 atm: V = \frac{RT}{P} = \frac{0.082057\times 298.15}{1} \approx 24.46\ \text{L per mol} = 0.02446\ \text{m}^3\text{ per mol}.
  • Henry’s law (gas solubility in water)
    • Solubility: [\text{gas}{(aq)}] = kH P_{\text{gas}}.
    • For CO₂ at 25 °C, a typical value used is around 3.3×10^{-2} M atm^{-1} (the exact value depends on the source).
  • Water autoionization (Kw) and pH
    • Water dissociation: K_w = [H^+][OH^-] = 1.0\times 10^{-14} \text{ at 25 °C}.
    • Neutral water: [H^+] = [OH^-] = 1.0\times 10^{-7}\ \text{M}, \quad \text{pH} = -\log_{10}[H^+] = 7.0.
  • Acid-base carbonate system in natural waters
    • CO₂(aq) hydration and protonation: CO₂(aq) + H₂O ⇌ H⁺ + HCO₃⁻; equilibrium constant K1 = \frac{[H^+][HCO3^-]}{[CO_2(aq)]} = 4.47\times 10^{-7}.
    • HCO₃⁻ ⇌ H⁺ + CO₃^{2-}; equilibrium constant K2 = \frac{[H^+][CO3^{2-}]}{[HCO_3^-]} = 4.68\times 10^{-11}.
    • In many cases, [CO₃^{2-}] is negligible relative to [HCO₃⁻] for near-neutral to acidic waters.
  • Rainwater pH: an example calculation
    • With only CO₂ and water, electroneutrality and Henry’s law yield a modest acidity.
    • If atmospheric CO₂ is 360 ppm at 25 °C, solving with K₁, K₂, and Henry’s law gives a rainwater pH ≈ 5.63 (example result from the lecture).
  • Ocean acidification (qualitative)
    • Ocean pH currently about 8.1; pre-industrial ~8.2; projected to around 7.8 by end of century under business-as-usual CO₂ emissions.
    • Lower pH reduces carbonate ion (CO₃^{2-}) availability, harming calcifying organisms and shells (CaCO₃).
  • Acid rain (qualitative)
    • Emissions of SO₂ and NOx form sulfuric and nitric acids in rain, lowering pH, sometimes to as low as ~1.8 in extreme cases.
    • Ecological impacts include biodiversity loss, shell dissolution, and aquatic dead zones; mitigation includes reducing SO₂/NOx emissions and using cleaner fuels.
  • Ammonia in water (base chemistry example)
    • NH₃(aq) + H₂O ⇌ NH₄⁺ + OH⁻; base dissociation constant Kb = \frac{[NH4^+][OH^-]}{[NH_3]} \approx 1.7\times 10^{-5}.
    • Henry’s law for NH₃: [NH3] = kH P{NH3} with k_H \approx 62\ \text{M atm}^{-1}.
    • Given a NH₃ air concentration of 50 ppm (≈ 50×10^{-6} atm), compute [NH_3] \approx 62 \times 50\times 10^{-6} \approx 3.1\times 10^{-3}\ \text{M}.
    • Then [NH4^+][OH^-] = Kb [NH_3] \approx (1.7\times 10^{-5})(3.1\times 10^{-3}) \approx 5.27\times 10^{-8}\ \text{M}^2.
    • Using electroneutrality and Kw, solve for [H⁺] to obtain pH ≈ 10.4 (alkaline solution).
  • Practical takeaways
    • The carbonate system largely controls pH in natural waters; CO₂ dissolution drives acidity, while biological and geological buffering can mitigate extremes.
    • Acid rain and ocean uptake of CO₂ are major environmental concerns with real-world policy implications.
    • The mathematics involves careful use of mass balances, first-order decay, and equilibria constants, plus correct unit handling.

Worked examples and balancing reminders

  • Balancing a methane combustion example to illustrate stoichiometry:
    • Unbalanced: CH₄ + O₂ → CO₂ + H₂O
    • Balanced form: CH₄ + 2 O₂ → CO₂ + 2 H₂O
  • Butane combustion (C₄H₁₀) example (derived in the lecture)
    • Balanced reaction (with integers): 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
    • Molar mass calculations:
    • Butane (C₄H₁₀): 4×12 + 10×1 = 58 g/mol
    • Carbon dioxide (CO₂): 12 + 2×16 = 44 g/mol
    • Convert 100 g butane to moles: n{C4H_{10}} = \frac{100}{58} \approx 1.724\ \text{mol}.
    • From stoichiometry, moles CO₂ produced: n{CO2} = 4 \times n{C4H_{10}} \approx 6.896\ \text{mol}.
    • Mass of CO₂ produced: m{CO2} = n{CO2} \times 44\ \text{g/mol} \approx 303\ \text{g}.
  • The ideal gas relationships and environmental applications were used to illustrate how gases behave, how to convert between units, and how to apply these concepts to air quality standards (e.g., SO₂ standard in mg/m³).

Connections to real-world environmental engineering and ethics

  • Mass balance methods underpin wastewater treatment design, pollutant tracking, and system optimization.
  • Understanding acid-base and carbonate equilibria helps in predicting rainwater acidity, lake recovery, and ocean buffering capacity.
  • Ocean acidification and acid rain exemplify how anthropogenic emissions affect biogeochemical cycles, with cascading effects on ecosystems, fisheries, and cultural heritage (e.g., limestone deterioration).
  • Policy implications include reducing CO₂ emissions, regulating SO₂/NOx emissions, adopting cleaner energy, and remediation strategies like liming acidic lakes.

Summary of key formulas to remember

  • CSTR steady-state with decay:
    C{\text{out}} = \frac{Q{\text{in}} C{\text{in}}}{Q{\text{out}} + kV}.

  • CSTR transient after a step-change in input for a non-steady-state, non-conservative pollutant:
    C(t) = C{\infty} + (C0 - C{\infty}) \exp\left[-\left(\frac{Q{\text{out}}}{V} + k\right) t\right].
    C{\infty} = \frac{Q{\text{in}} C{\text{in, new}}}{Q{\text{out}} + kV},\quad C0 = C{\text{out}}(t=0).

  • PFR decay:
    C(t) = C_0 e^{-kt}.

  • k from half-life (first-order):
    k = -\frac{1}{t} \ln\left(\frac{Ct}{C0}\right).

  • Ideal gas volume per mole (STP and 25 °C):
    V{\text{STP}} = 22.4\ \text{L mol}^{-1} = 0.0224\ \text{m}^{3}\text{ mol}^{-1}, V{25^{\circ}\text{C}} = \frac{RT}{P} \approx 24.46\ \text{L mol}^{-1} = 0.02446\ \text{m}^{3}\text{ mol}^{-1}.

  • Henry’s law (gas in water):
    [\text{gas}{(aq)}] = kH P_{\text{gas}}.

  • Water autowionization (Kw):
    K_w = [H^+][OH^-] = 1.0 \times 10^{-14}.

  • Carbonate system (aqueous CO₂ dissolution):
    K1 = \frac{[H^+][HCO3^-]}{[CO2(aq)]} = 4.47\times 10^{-7}, K2 = \frac{[H^+][CO3^{2-}]}{[HCO3^-]} = 4.68\times 10^{-11}.

  • Rainwater pH example (CO₂-driven): approximate result pH ≈ 5.6–5.7 under 360 ppm CO₂ at 25 °C, illustrating natural acidity.

  • Ammonia equilibrium in water (base):
    NH3 + H2O \rightleftharpoons NH4^+ + OH^-, Kb = \frac{[NH4^+][OH^-]}{[NH3]} \approx 1.7\times 10^{-5}.

  • Ammonia solubility (Henry’s law):
    [NH3] = kH P{NH3},\quad k_H \approx 62\ \text{M atm}^{-1}.$$