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.