1 / 14

Gallery Walk problems

This article discusses several key thermodynamic calculations related to various chemical reactions, including the calculation of Gibbs free energy (ΔG) under both standard and non-standard conditions. It examines reactions involving iodine, chlorine, carbon monoxide, and precipitate formation using silver chloride, as well as addresses the solubility of magnesium hydroxide. Key principles of Ksp, and how to determine the mass of precipitates and the necessary amounts of KOH to achieve complete precipitation of metal ions in solution, are also covered.

tibor
Télécharger la présentation

Gallery Walk problems

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. Gallery Walk problems

  2. Consider the following reaction: I2 (g) + Cl2 (g)  2 ICl (g) Kp = 81.9 @ 25°C Calculate Grxn @ 25°C under the following conditions: Standard conditions Equilibrium PICl = 2.55 atm, PI2 = 0.325 atm, PCl2 = 0.221 atm G° = -RTlnK G° = - 8.314 J/molK*298 K * ln (81.9) G° = -1.09x104 J/mol b) G = 0 c) G = G° + RT ln Q G = -1.09x104 J/mol + 8.314*298* ln [(2.55)2/(0.325*0.221)] G = 249 J/mol

  3. Consider the following reaction: CO(g) + Cl2(g)  2 COCl2 (g) Calculate G for this reaction at 25°C if PCO = 0.112 atm, PCCl4 = 0.174 atm, PCOCl2 = 0.774 atm. G° = 2*Gf°(COCl2) – [Gf°(CO) + Gf°(Cl2)] G° = 2*(-204.9 kJ/mol) – [-137.2 kJ/mol + 0 kJ/mol] G° = -272.6 kJ/mol G = G° + RTlnQ G = -272600 J/mol + 8.314 J/molK*298*ln(30.7) G = -264,110 J/mol

  4. What mass of precipitate will form upon mixing 175.0 mL of a 0.0055 M KCl solution with 145.0 mL of a 0.0015 M AgNO3 solution? 0.0055 M KCl * (175.0 mL/320.0 mL) = 0.00301 M Cl- 0.0015 M AgNO3 * (145.0 mL/320.0 mL) = 0.000680 M Ag+ Ksp(AgCl) = 1.8x10-10

  5. Previous problem continued AgCl(s) = Ag+ (aq) + NO3-(aq)

  6. Cont’d 1.8x10-10 = (x)(0.00233+x) Assume x<<0.00233 1.8x10-10 = x(0.00233) X = 7.7253x10-8 M 0.000680 M AgCl * 0.320 L = 2.176x10-4 mol AgCl 2.176x10-4 mol AgCl * 143.32 g/mol = 0.0312 g AgCl

  7. What is the solubility (in g/mL) of magnesium hydroxide in a solution buffered at pH = 10? Ksp = 6.3x10-10 Ksp = [Mg2+][OH-]2 pOH = 14 – pH = 4 [OH] = 10-4 6.3x10-10 = [Mg2+][10-4]2 [Mg2+] = 0.063 S = 0.063 mol/L *58.31 g/mol = 3.67 g/L * 1 L/1000 mL = 0.00367 g/mL

  8. Calculate K at 25 C for the following reaction: 2 CO (g) + O2 (g)  2 CO2 (g) G° = -514.4 kJ/mol G° = - RT ln K -514400 J/mol = - 8.314 J/mol K * 298 K ln K K = 1.23

  9. At what temperatures is the following reaction spontaneous: CaCO3 (s) = CaO (s) + CO2 (g) Hrxn° = 178 kJ/mol Srxn° = 159.6 J/mol K G= Hrxn° - T Srxn° 0 = 178000J/mol – T *159.6 J/mol K T = 1115 K T>1115 K

  10. The solubility of CuCl is 3.91 mg per 100.0 mL. What is the Ksp for CuCl? Ksp = [Cu+][Cl-] 3.91 mg/100.0 mL = .0391 g/L 0.0391 g/L * 1 mol/99 g = 3.95x10-2 M Ksp = (3.95x10-2 M)2 Ksp = 1.56x10-7

  11. I want to precipitate the metal ions from 100.0 mL of a solution that is 0.100 M in Ca2+, Mg2+, and Fe2+. How much KOH do I need to add to start the precipitation of each metal? How much total KOH would I need to add to precipitate all of the metal ions? Three relevant reactions: Ca(OH)2 (s) = Ca2+ (aq) + 2 OH-(aq) Ksp = 6.5x10-6 Fe(OH)2 (s) = Fe2+ (aq) + 2 OH-(aq) Ksp = 4.1x10-15 Mg(OH)2 (s) = Mg2+ (aq) + 2 OH-(aq) Ksp = 6.3x10-10

  12. Cont’d Fe(OH)2 (s) = Fe2+ (aq) + 2 OH-(aq) Ksp = 4.1x10-15 = (0.100 M) [OH-]2 [OH-] = 2.025x10-7 M * 0.1 L * 56.1 g/mol = 1.14x10-6 g. Since the Ksp is so different, we can assume that the reactions are separate. To TOTALLY precipitate the Fe(OH)2 requires: 0.100 M * 0.1 L * 2 OH-/1 Fe * 56.1 g/mol = 1.12 g Then on to Mg.

  13. Cont’d Mg(OH)2 (s) = Mg2+ (aq) + 2 OH-(aq) Ksp = 6.3x10-10 = (0.100 M) [OH-]2 [OH-] = 7.94x10-5 M * 0.1 L * 56.1 g/mol = 4.45x10-4 g + 1.12 g to precipitate Fe. Since the Ksp is so different, we can assume that the reactions are separate. To TOTALLY precipitate the Mg(OH)2 requires: 0.100 M * 0.1 L * 2 OH-/1 Mg * 56.1 g/mol = 1.12 g + 1.12 g required for the Fe = 2.24 g Then on to Ca

  14. Cont’d Ca(OH)2 (s) = Ca2+ (aq) + 2 OH-(aq) Ksp = 6.5x10-6 = (0.100 M) [OH-]2 [OH-] = 8.06x10-3 M * 0.1 L * 56.1 g/mol = 4.52x10-2 g + 2.24 g to precipitate Fe and Mg.. Since the Ksp is so different, we can assume that the reactions are separate. To TOTALLY precipitate the Ca(OH)2 requires: 0.100 M * 0.1 L * 2 OH-/1 Ca * 56.1 g/mol = 1.12 g + 1.12 g + 1.12 g = Done.

More Related