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Gallery Walk problems

Gallery Walk problems. Consider the following reaction: I 2 (g) + Cl 2 (g)  2 ICl (g) K p = 81.9 @ 25°C Calculate G rxn @ 25°C under the following conditions: Standard conditions Equilibrium P ICl = 2.55 atm, P I2 = 0.325 atm, PCl 2 = 0.221 atm. G° = -RTlnK

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Gallery Walk problems

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  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.

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