1. Gallery Walk problems

2. 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: a)Standard conditions b)Equilibrium c)P ICl = 2.55 atm, P I2 = 0.325 atm, PCl 2 = 0.221 atm a)  G° = -RTlnK  G° = - 8.314 J/mol  K*298 K * ln (81.9)  G° = -1.09x10 4 J/mol b)  G = 0 c)  G =  G° + RT ln Q  G = -1.09x10 4 J/mol + 8.314*298* ln [(2.55) 2 /(0.325*0.221)]  G = 249 J/mol

3. Consider the following reaction: CO(g) + Cl 2 (g)  2 COCl 2 (g) Calculate  G for this reaction at 25°C if P CO = 0.112 atm, P CCl4 = 0.174 atm, P COCl2 = 0.774 atm.  G° = 2*  G f °(COCl 2 ) – [  G f °(CO) +  G f °(Cl 2 )]  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 AgNO 3 solution? 0.0055 M KCl * (175.0 mL/320.0 mL) = 0.00301 M Cl - 0.0015 M AgNO 3 * (145.0 mL/320.0 mL) = 0.000680 M Ag + K sp (AgCl) = 1.8x10 -10

5. Previous problem continued AgCl(s) = Ag + (aq) + NO 3 - (aq) I 00.000680 M0.00301 M C +0.000680-0.000680 I 0.00068000.00233 M C -x+x E 0.000680x0.00233+x

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? K sp = 6.3x10 -10 K sp = [Mg 2+ ][OH-] 2 pOH = 14 – pH = 4 [OH] = 10 -4 6.3x10 -10 = [Mg 2+ ][10 -4 ] 2 [Mg 2+ ] = 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) + O 2 (g)  2 CO 2 (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: CaCO 3 (s) = CaO (s) + CO 2 (g)  H rxn ° = 178 kJ/mol  S rxn ° = 159.6 J/mol K  G=  H rxn ° - T  S rxn ° 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 K sp for CuCl? K sp = [Cu+][Cl-] 3.91 mg/100.0 mL =.0391 g/L 0.0391 g/L * 1 mol/99 g = 3.95x10 -2 M K sp = (3.95x10 -2 M) 2 K sp = 1.56x10 -7

11. I want to precipitate the metal ions from 100.0 mL of a solution that is 0.100 M in Ca 2+, Mg 2+, and Fe 2+. 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) = Ca 2+ (aq) + 2 OH - (aq) K sp = 6.5x10 -6 Fe(OH) 2 (s) = Fe 2+ (aq) + 2 OH - (aq) K sp = 4.1x10 -15 Mg(OH) 2 (s) = Mg 2+ (aq) + 2 OH - (aq) K sp = 6.3x10 -10

12. Cont’d Fe(OH) 2 (s) = Fe 2+ (aq) + 2 OH - (aq) K sp = 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 K sp 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) = Mg 2+ (aq) + 2 OH - (aq) K sp = 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 K sp 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) = Ca 2+ (aq) + 2 OH - (aq) K sp = 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 K sp 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.