1 / 42

Methods of Green Chemistry in Environmental Technology

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

efia
Download Presentation

Methods of Green Chemistry in Environmental Technology

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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:

More Related