gallery walk problems. consider the following reaction: i 2 (g) + cl 2 (g) 2 icl (g) k p = 81.9 @...
Post on 04-Jan-2016
216 Views
Preview:
TRANSCRIPT
Gallery Walk problems
Consider the following reaction:
I2 (g) + Cl2 (g) 2 ICl (g)
Kp = 81.9 @ 25°C
Calculate Grxn @ 25°C under the following conditions:
a) Standard conditionsb) Equilibrium
c) PICl = 2.55 atm, PI2 = 0.325 atm, PCl2 = 0.221 atm
a) G° = -RTlnK G° = - 8.314 J/molK*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
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
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
Previous problem continued
AgCl(s) = Ag+ (aq) + NO3-(aq)
I 0 0.000680 M 0.00301 M
C +0.000680 -0.000680 -0.000680
I 0.000680 0 0.00233 M
C -x +x +x
E 0.000680 x 0.00233+x
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 AgCl2.176x10-4 mol AgCl * 143.32 g/mol = 0.0312 g
AgCl
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
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
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
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
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
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.
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
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.
top related