CEE 370 Environmental Engineering Principles Lecture #7 - PowerPoint PPT Presentation

Print version Updated: 18 September 2019 CEE 370 Environmental Engineering Principles Lecture #7 Environmental Chemistry V: Thermodynamics, Henry’s Law, Acids-bases II Reading: Mihelcic & Zimmerman, Chapter 3 Davis & Masten, Chapter 2 Mihelcic, Chapt 3 David Reckhow CEE 370 L#7 1

Henry’s Law Henry's Law states that the amount of a gas that dissolves into a liquid is proportional to the partial pressure that gas exerts on the surface of the liquid. In equation form, that is: C = K p A H A where, C A = concentration of A, [mol/L] or [mg/L] K H = equilibrium constant (often called Henry's Law constant), [mol/L-atm] or [mg/L-atm] p A = partial pressure of A, [atm] 2 CEE 370 L#7 David Reckhow

Henry’s Law Constants Reaction Name K h , mol/L-atm pK h = -log K h 3.41 x 10 -2 CO 2 (g) _ CO 2 (aq) Carbon 1.47 dioxide NH 3 (g) _ NH 3 (aq) Ammonia 57.6 -1.76 1.02 x 10 -1 H 2 S(g) _ H 2 S(aq) Hydrogen 0.99 sulfide 1.50 x 10 -3 CH 4 (g) _ CH 4 (aq) Methane 2.82 1.26 x 10 -3 O 2 (g) _ O 2 (aq) Oxygen 2.90 3 CEE 370 L#7 David Reckhow

Example: Solubility of O 2 in Water  Background  Although the atmosphere we breathe is comprised of approximately 20.9 percent oxygen, oxygen is only slightly soluble in water. In addition, the solubility decreases as the temperature increases. Thus, oxygen availability to aquatic life decreases during the summer months when the biological processes which consume oxygen are most active.  Summer water temperatures of 25 to 30 ° C are typical for many surface waters in the U.S. Henry's Law constant for oxygen in water is 61.2 mg/L-atm at 5 ° C and 40.2 mg/L-atm at 25 ° C.  What is the solubility of oxygen at 5 ° C and at 25 ° C? Example 4.1 from Ray 4 CEE 370 L#7 David Reckhow

Solution O 2 Solubility Ex. At 5 0 C the solubility is: mg ° C (5 C) = K P = 61.2 L-atm x 0.209 atm O H,O O 2 2 2 C (5 C) = 12.8 mg ° O L 2 At 25 0 C the solubility is: mg ° C (25 C) = K P = 40.2 L-atm x 0.209 atm H,O O O 2 2 2 C (25 C) = 8.40 mg ° O L 2 5 CEE 370 L#7 David Reckhow

Air Stripping Tower Air out with Carbon Dioxide Water in with Carbon Air In Water Out 6 CEE 370 L#7 David Reckhow

Air Stripping Example An air stripping tower, similar to that shown in the previous slide , is to be used to remove dissolved carbon dioxide gas from a groundwater supply. If the tower lowers the level to twice the equilibrium concentration, what amount of dissolved gas will remain in the water after treatment? The partial pressure is about 350 ppm or 0.00035. The log of 0.00035 is about -3.5. Therefore the partial pressure of carbon dioxide in the atmosphere is 10 -3.5 atm. Example 4.2 from Ray 7 CEE 370 L#7 David Reckhow

Solution to Air Stripping Ex. The first step is to determine Henry's Law constant for carbon dioxide. From the table it is 10 -1.5 . The equilibrium solubility is then: mole mole − = 5 = = - 1 .5 - 3 .5 atm 10 p C K 10 10 L H, CO CO CO L- atm 2 2 2 3 = 10 M = 10 mole x 44 g mole x 10 mg -5 -5 C CO L g 2 = 0.44 mg/ L C At equilibrium: CO 2 = 0. 88 mg/L After treatment: C CO 2 8 CEE 370 L#7 David Reckhow

Ideal Gas Law The Ideal Gas Law states that the product of the absolute pressure and the volume is proportional to the product of the mass and the absolute temperature. In equation form this is usually written: PV = nRT where, P = absolute pressure, [atm] V = volume, [L] n = mass, [mol] T = absolute temperature, [K] R = proportionality constant or ideal gas constant, [0.0821 L-atm/K-mol] 9 CEE 370 L#7 David Reckhow

Ideal Gas Law Example Anaerobic microorganisms metabolize organic matter to carbon dioxide and methane gas. Estimate the volume of gas produced (at atmospheric pressure and 25 ° C) from the anaerobic decomposition of one mole of glucose. The reaction is: C 6 H 12 O 6 → 3CH 4 + 3CO 2 Example 4.3 from Ray 10 CEE 370 L#7 David Reckhow

Solution to Ideal Gas Law Ex. Each mole of glucose produces three moles of methane and three moles of carbon dioxide gases, a total of six moles. The total volume is then: (6 mol)(0.0821 L-atm K- mol)(298 K) V = nRT = P (1 atm) V = 147 L 11 CEE 370 L#7 David Reckhow

Dalton’s Law of Partial Pressure Dalton's Law of partial pressures states that the total pressure of a mixture of several gases is the sum of the partial pressures of the individual gases. In equation form this is simply: n ∑ P = P i t i=1 where, P t = total pressure of the gases, [atm] P i = pressure of the ith gas, [atm] 12 CEE 370 L#7 David Reckhow

Anaerobic Digester Example Anaerobic digesters are commonly used in wastewater treatment. The biological process produces both carbon dioxide and methane gases. A laboratory worker plans to make a "synthetic" digester gas. There is currently 2 L of methane gas at 1.5 atm and 1 L of carbon dioxide gas at 1 atm in the lab. If these two samples are mixed in a 4 L tank, what will be the partial pressures of the individual gases? The total pressure? Example 4.4 from Ray 13 CEE 370 L#7 David Reckhow

Solution to Anaerobic Digester Ex. First, we must find the partial pressures of the individual gases using the ideal gas law:   V 1 P V = nRT = P V or   P = P 1 1 2 2 2 1   V 2   P = 1.5 atm 2 L   For methane gas 4 L = 0.75 atm   2   P = 1 atm 1 L   4 L = 0.25 atm For carbon dioxide gas:   2 P = P + P = 1 atm And the total is: t CH CO 4 2 14 CEE 370 L#7 David Reckhow

Equilibrium Reactions ↔ aA + bB pP + rR c d = {C } {D } K thermodynamics eq a b {A } {B } a, b, p, r = stoichiometric coefficients of the respective reactants {A}, {B}, {P}, {R} = activity of the reactants and products 15 CEE 370 L#7 David Reckhow

Chemical Activity γ {A} = [A] A {A} = activity of species A, [mol/liter] [A] = concentration of species A, [mol/liter] γ A = activity coefficient of A, unitless γ is dependent on ionic strength -- the concentration of ions in solution. For dilute aqueous solutions γ is near unity. Thus, the activity and concentration are approximately the same. For most freshwater systems, γ may be neglected. 16 CEE 370 L#7 David Reckhow

Activity conventions  the activity of the solvent , water is set equal to unity  the activity of a solid in equilibrium with a solution is unity  the activity of a gas is the partial pressure the gas exerts on the liquid surface  the activity of a solute is related to the concentration by {A} = γ A [A], where γ A is the activity coefficient of species A 17 CEE 370 L#7 David Reckhow

Auto-dissociation of Water ↔ + - H O H + OH 2 + - K = [ H ][OH ] + - = [ H ][OH ] w [ H O] 2 At 25 ° C the value of K w is 10 -14 18 CEE 370 L#7 David Reckhow

Example 4.12 Find the pH and pOH of water at 25 ° C if the concentration of H + ions is 10 -5 M. Solution + -5 pH = -log{H } = -log(10 ) = 5 -14 {OH } = K {H } = 10 w - -9 = 10 + -5 10 - -9 pOH = -log{OH } = -log(10 ) = 9 19 CEE 370 L#7 David Reckhow

Acids & Bases  Which is the strongest acid A. H 3 PO 4 B. H 2 CO 3 C. HNO 3 D. CH 3 COOH E. HF 20 CEE 370 L#7 David Reckhow

Acid:Base Equilibria Monoprotic Acid + - K = {H }{ A } ↔ + - HA H + A A {HA} Diprotic Acid + - K 1 = {H }{HA } + - H A = H + HA A {H A} 2 2 + 2- K 2 = {H }{ A } ↔ - + 2- HA H + A A - {HA } 21 CEE 370 L#7 David Reckhow

Similar to Table 3.5 in M&Z Acidity Constants Reaction Name K a pK a = ─ log K a + + Cl ─ HCl = H Hydrochloric 1000 -3 + + HSO 4 - H 2 SO 4 = H Sulfuric, H1 1000 -3 + + NO 3 - HNO 3 = H Nitric ~1 ~0 - = H + + SO 4 - -2 HSO 4 Sulfuric, H2 1 x 10 2 + + H 2 PO 4 - -3 H 3 PO 4 = H Phosphoric, H1 7.94 x 10 2.1 + + Ac - -5 HAc = H Acetic 2.00 x 10 4.7 + + HCO 3 - -7 H 2 CO 3 = H Carbonic, H1 5.01 x 10 6.3 + + HS - -8 H 2 S = H Hydrosulfuric, H1 7.94 x 10 7.1 - = H + + HPO 4 -8 H 2 PO 4 ─ 2 Phosphoric, H2 6.31 x 10 7.2 + + OCl - -8 HOCl = H Hypochlorous 3.16 x 10 7.5 + = H + + NH 3 -10 NH 4 Ammonium 5.01 x 10 9.3 - = H + + CO 3 -11 ─ 2 HCO 3 Carbonic, H2 5.01 x 10 10.3 + + PO 4 -13 HPO 4 ─ 2 = H ─ 3 Phosphoric, H3 5.01 x 10 12.3 22 CEE 370 L#7 David Reckhow