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Methods of Green Chemistry in Environmental Technology

Methods of Green Chemistry in Environmental Technology. Exercises I Eveliina Repo. Membranes. Osmotic pressure.

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Methods of Green Chemistry in Environmental Technology

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  1. Methods of Green Chemistry in Environmental Technology Exercises I Eveliina Repo

  2. Membranes

  3. Osmotic pressure • Osmotic pressure arises when two solutions of different concentrations, or a pure solvent and a solution, are separated by a semipermeable membrane. Molecules such as solvent molecules that can pass through the membrane will migrate from the side of higher concentration to the side of lower concentration in a process known as osmosis. • The pressure required to stop osmosis is called the osmotic pressure. • In dilute solutions, osmotic pressure is directly proportional to the molarity of the solution and its temperature in Kelvin

  4. Osmotic pressure

  5. Osmotic pressure • Calculate the osmotic pressure exhibited • by a 0.42M KOH solution at 30oC • by a 0.35M Na2SO4 solution at 25oC

  6. Osmotic pressure R = 8.314 J K-1 mol-1 = 0.0821 L atmK-1 mol-1 T = 303 K C (K+) = 0.42 mol/L C (OH-) = 0.42 mol/L C (solute) = 0.84 mol/L

  7. Osmotic pressure R = 8.314 J K-1 mol-1 = 0.0821 L atmK-1 mol-1 T = 298 K C (Na+) = 0.7 mol/L C (SO4) = 0.35 mol/L C (solute) = 1.05 mol/L

  8. Osmoticpressure • A solution of 20.0 g of polyisobutylene in 1.00 dm3 of benzene was placed in an osmometer, at 25°C. After equilibrium had been obtained, the height h was found to be 24.45 mm of benzene. Find the average molar mass of the polymer. The density of the solution is 0.879 g cm–3 h solvent solution membrane

  9. Osmoticpressure

  10. Osmotic pressure • For 1 dm3 solution: • Mass of polyisobutylene was 20 g when its molar mass is:

  11. Water (solution) flux Water diffusion can be expressed by a classical solution-diffusion model Dw = diffusivity in membrane, cm2/s or m2/s Cw = average water conc. in membrane, g/cm3 (~ 0.2) Vw = partial molar volume of water, cm3/g ΔP = pressure difference R = gas constant T = temperature Δπ = osmotic pressure difference z = membrane thickness Aw = permeability constant (at particular temperature)

  12. Salt (solute) flux Salt flux in given as: Ds = diffusivity of solute in membrane, cm2/sor m2/s Ks = distribution coefficient C1 and C2 = feed side and permeate side solute concentration g /L z = membrane thickness As = solute permeability constant , m/s

  13. Water / salt flux • A reverse osmosis unit is to deminerialize 750 000 L/day effluent. Pertinent data are: permeability coefficient: 0.2 L/(m2 day kPa) at 25 oC, pressure difference between the feed and product water: 2500 kPa, osmotic pressure difference between the feed and product water: 300 kPa. Determine the membrane area required if the lowest operation temperature is 10 oC. • Membrane area correction (AT/A25) should be considered as the following: For 10 oC,1.58; 15oC, 1.34; 20oC, 1.15; 25oC, 1.00; 30oC, 0.84

  14. Water / salt flux • A water flux is: (A10/A25) • Area required:

  15. Water / salt flux • A 1 m2 reverse osmosis membrane is being used to filter salt from a 0.1 M NaCl solution (measured osmotic pressure difference: 4.1 atm) using a 39.8% acetyl cellulose acetate membrane (Dw = 1.6*10-6 cm2/s, Cw = 0.0089 mol/cm3). (a) What is the maximum water flux through this membrane at a pressure DP = 27.2 atm and 25 oC. The membrane is approximately 40 mm thick? (b) How much membrane area would be required to produce a liter of water in 1 hour? Vw = 18 cm3/mol.

  16. Water / salt flux

  17. Water / salt flux Desired amount of water: 1 L/h = 1000 cm3 / 3600 s = 0.27778 cm3/s Volume flux through 1 m2 of membrane: 0.0109 cm3/s m2

  18. RO unit • Design a membrane filtration system (the amount of elements and units and flow rates of units) when the requirement of reverse osmosis water is 3785 m3/d. The flux in membrane element is 610 L/m2 4-2-1–system: Stage 1: 4 pressure vessels, Stage 2: 2 pressure vessels, Stage 3: 1 pressure vessel. There are seven elements in one pressure vessel and the area of one element is 33 m2. Yield of permeate is 75%. Concentrate Permeate

  19. RO unit Water filtered by one element: The amount of the elements = Filtrated water by 49 elements: The amount of membrane units needed: → 4 units • The amount of elements:

  20. RO unit

  21. Electrochemistry

  22. Balancing electrochemical equations • Half reactions are: • Balance oxygen with water, hydrogen with protons and charge with electrons.

  23. Balancing electrochemical equations • Summarize the half reactions: × 2 × 5

  24. Balancing electrochemical equations • Balance the following equation: Cr2O72- (aq) + Cl- (aq) + H+ (aq)  Cr3+ (aq) + Cl2 (g) + H2O (l)

  25. Balancing electrochemical equations Half reactions are: Cr2O72- 2 Cr3+ (reduction) and 2 Cl- Cl2 (oxidation) Oxygen is balanced with water: Cr2O72- 2 Cr3+ + 7 H2O • Hydrogen is balanced with protons: • Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O

  26. Balancing electrochemical equations Charge is balanced with electrons: Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O and 2 Cl- Cl2 + 2 e- Both sides should contain same amount of electrons: Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O 2 Cl- Cl2 + 2 e- × 3 Cr2O72- + 6 Cl- + 14 H+ 2 Cr3+ + 3 Cl2 + 7 H2O

  27. Electrochemical cell Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) Reduction: Cu2+ + 2 e- Cu Eored = 0.34 V Oxidation: Zn  Zn2+ + 2 e-Eoox = -(-0.76 V) = 0.76 V Cu2+ + 2 e- Cu Eored = 0.34 V Zn  Zn2+ + 2 e-Eoox = 0.76 V Zn + Cu2+ Zn2+ + Cu Oxidation - Losing electrons Reduction - Gaining electrons Eocell = Eored + Eoox = 1.10 V PositiveEocell means spontaneous reaction

  28. E0 is for the reaction as written • The more positive E0 the greater the tendency for the substance to be reduced • The half-cell reactions are reversible • The sign of E0changes when the reaction is reversed • Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0 19.3

  29. Electrochemical cell • Calculate the cell potentials for the following reactions: • (a) 2 Ag+ + Sn → 2 Ag + Sn2+ • (b) Au + NO3- + 4 H+ → Au3+ + NO + 2H2O • (c) 2 MnO4- + 5 Fe +16 H+→ 5 Fe2+ + 2 Mn2+ + 8 H2O

  30. Electrochemical cell 2 Ag+ + 2 e- → 2 Ag Eored = + 0.80 V Sn → Sn2+ + 2 e- Eoox= - (-0.14) V 2 Ag+ + Sn → 2 Ag + Sn2+ Eocell = Eored + Eoox = 0.80 V + 0.14 V = 0.94 V NO3- + 4 H+ + 3 e- → NO + 2H2O Eored = + 0.96 V Au → Au3+ + 3 e- Eoox =- (+ 1.50) V Au + NO3- + 4 H+ → Au3+ + NO + 2H2O Eocell = Eored + Eoox = 0.96 V - 1.50 V = -0.54 V Reaction will not happen i.e. gold cannot be dissolved in nitric acid

  31. Electrochemical cell 2 MnO4-+ 16 H+ +10 e- → 2 Mn2+ + 8 H2O Eored = 1.51 V 5 Fe → 5 Fe2+ + 10 e- Eoox =- (-0.44) V 2 MnO4- + 5 Fe +16 H+→ 5 Fe2+ + 2 Mn2+ + 8 H2O Eocell = Eored + Eoox = 1.51 V + 0.44 V = 1.95 V

  32. Electrochemical cell • Gibbs energy and Nernst equation T = 25 oC

  33. Electrochemical cell • Calculate the standard Gibbs free energy for the following reactions: • (a) 2 Ag+ + Sn → 2 Ag + Sn2+ • (b) Au + NO3- + 4 H+ → Au3+ + NO + 2H2O • (c) 2 MnO4- + 5 Fe +16 H+→ 5 Fe2+ + 2 Mn2+ + 8 H2O

  34. Electrochemical cell • (a) 2 Ag+ + Sn → 2 Ag + Sn2+Eocell = 0.94 V = 0.94 J/C • (b) Au + NO3- + 4 H+ → Au3+ + NO + 2H2O Eocell = -0.54 V = -0.54 J/C

  35. Electrochemical cell • (c) 2 MnO4- + 5 Fe +16 H+→ 5 Fe2+ + 2 Mn2+ + 8 H2O • Eocell = 1.95 V = 1.95 J/C

  36. Electrochemical cell Calculate the cell potential at 25 oC for the following reaction: 2 Al + 3 Mn2+→ 2 Al3+ + 3 MnEocell= 0.48 V [Mn2+] = 0.50 M [Al3+] = 1.50 M From half reactions: n = 6 2 Al → 2 Al3+ + 6 e- 3 Mn2+ + 6 e- → 3 Mn2+

  37. Electrochemical cell Calculate the cell potential at 25 oC for the following system: VO2+ + 2 H+ + e- → VO2++ H2O Eo = 1.00 V Zn2+ + 2 e- → ZnEo= -0.76 V [VO2+] = 2.0 M [H+] = 0.50 M [VO2+] = 0.01 M [Zn2+] = 0.1 M

  38. Electrochemical cell 2 VO2+ + 4 H+ + 2 e- → 2 VO2++ 2 H2O Eored = 1.00 V Zn→ Zn2+ + 2 e-Eoox= -(-0.76 V) 2 VO2+ + 4 H+ + Zn → 2 VO2++ 2 H2O + Zn2+ Eocell= 1.76 V

  39. Electrochemical cell • Electrolysis: forcing a current through a cell to produce a chemical change for which the cell potential is negative Zn2+ + Cu → Zn + Cu2+ Eocell= -1.10 V

  40. Electrochemical cell • How much Zn2+ can be removed from the wastewater using 10 A current for 30 min? 1 A = 1 C/s (a coulomb of charge per second) Coulombs of charge = 10 C/s × 30 × 60 s = 1.8 × 104 C 1 mole of electrons carries 96 485 coulombs Amount of electrons = 1.8 × 104 C × 1 mol e- / 96 485 C = 0.187 mol e-

  41. Electrochemical cell Each Zn2+ ion requires two electrons to become a Zn atom. Thus each mole of electrons produces ½ mole of Zn: Amount of Zn = 1/2 × 0.187 mol = 0.0935 mol In grams: m = nM= 0.0935 mol × 65.39 g/mol = 6.11 g

  42. Electrochemical cell • What is the time required to remove all Ag+ ions from the 0.5 L of solution with Ag+ concentration of 21 g/L using 5 A current? • Amount of Ag+: • Amount of electrons required: • Charge of these electrons: • Time required:

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