Transcript Gallery Walk problems

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 conditions
b) Equilibrium
c) PICl = 2.55 atm, PI2 = 0.325
atm, PCl2 = 0.221 atm
a) 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
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
0.0055 M KCl * (175.0
precipitate will form
mL/320.0 mL) =
upon mixing 175.0 mL
0.00301 M Clof a 0.0055 M KCl
solution with 145.0
0.0015 M AgNO3 *
mL of a 0.0015 M
(145.0 mL/320.0 mL)
AgNO3 solution?
= 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 AgCl
2.176x10-4 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:
Hrxn° = 178 kJ/mol
Srxn° = 159.6 J/mol K
CaCO3 (s) = CaO (s) + CO2 (g)
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.