CHEM 10123/10125, Quiz 5
March 28, 2012
Name_____________________
(please print)
Significant figures, phases, and correct units count, show charges as appropriate, and please box your answers!
1. (5 points) SHOW ALL WORK. Elemental sulfur exists in two forms; an orthorhombic form
(Sortho), which is stable below 95.3 °C, and a monoclinic form (Smono), which is stable above 95.3
°C. The conversion of orthorhombic sulfur to monoclinic sulfur is endothermic, with H =
0.401 kJ/mol at 1 atm.
a) What is S for the conversion of orthorhombic to monoclinic sulfur?
b) Which is the more highly ordered form of sulfur, orthorhombic or monoclinic?
[answers; a) At 95.3 °C = 368.45 K, G = 0 = H – TS
So, (0.410) = (368.45) S
S = 1.09 J/mol K,
b) orthorhombic is more ordered]
2. (8 points) Write the balanced chemical equation that corresponds to each of the following
constants. Pay attention to phases, arrows, charges, and stoichiometry!
a) Gf for K2SO4(s)
2 K(s) + S(s) + 2 O2(g)  K2SO4(s)
b) Kf for CCl4(l)
mistaken problem! Gave 2 pts to all for this one.
c) Ksp for CaCO3 (s)
CaCO3(s) ⇌ Ca2+(aq) + CO32–(aq)
d) Hvap for NH3
NH3(l) ⇌ NH3(g)
3. (8 points) SHOW ALL WORK. In a saturated solution of Ca3(PO4)2 at 25 °C, calculate a)
the molarity of each ion, and b) the mass that will dissolve in 0.100 L of water. (FW Ca3(PO4)2
= 310.18 g/mol, Ksp = 2.07 x 10–33) (assume no volume change on dissolution)
Answer:
Ca3(PO4)2(s) ⇌ 3 Ca2+(aq) + 2 PO43–(aq)
a) Ksp = [Ca2+]3[PO43–]2 = (3x)3(2x)2 = 2.07 x 10–33 = (27x3)(4x2) = 108x5
x5 = 1.91666 x 10–35, so x = 1.14 x 10–7 M
[Ca2+] = 3.42 x 10–7 M
[PO43–] = 2.28 x 10–7 M
b) 0.100 L H2O x (1.14 x 10–7 mol salt)/(1 L soln) x (310.18 g salt)/(1 mol salt) = 3.54 x 10–6 g
Selected Thermodynamic Data for Materials at 25 °C.
Substance
H°f (kJ/mol)
S° (J/mol K)
G°f (kJ/mol)
CaCO3(s)
CaO(s)
CCl4(l)
CO2(g)
CO(g)
-1207.6
-634.9
-95.7
-393.5
-110.5
91.7
38.1
309.9
213.8
197.7
-1129.1
-603.3
-53.6
-394.4
-137.2
Cu(s)
Cu2O(s)
H2(g)
HCl(g)
H2O(l)
H2O(g)
H2O2(l)
H2SO4(l)
KCl(s)
K2SO4(s)
N2(g)
NH3(g)
NO(g)
NO2(g)
O2(g)
SO3(g)
0
-168.6
0
-92.3
-285.8
-241.8
-187.8
-814
-436.5
-1437.8
0
-45.9
91.3
33.2
0
-395.7
33.2
93.1
130.7
186.9
70.0
188.7
109.6
156.9
82.6
175.6
191.6
192.8
210.8
240.1
205.2
256.8
0
-146.0
0
-95.3
-237.1
-228.6
-120.4
-690.0
-408.5
-1321.4
0
-16.4
87.6
51.3
0
-371.1
4. (9 points) SHOW ALL WORK. For each of the following reactions, calculate the appropriate
quantity, using tables in the back.
a. G°, in kJ/mol -72.6 kJ/mol
2 NO(g) + O2(g) ⇌ 2 NO2(g)
G = [(2)(–51.3)] – [2(87.6) + 0] = 102.6 – 175.2 = –72.6 kJ/mol
b. G°, in kJ/mol –32.8 kJ/mol,
and the thermodynamic equilibrium constant Kp, at 25 °C 6 x 105
N2(g) + 3 H2(g) ⇌ 2 NH3(g)
G = (2)(–16.4) – [(0) + (3)(0)] = –32.8 kJ/mol
G = – RTln K = – (8.314 x 10–3)(298)ln K
ln K = 13.2387
K = 5.62 x 105 ~ 1 sig fig, so 6 x 105
CHEM 10123/10125, Quiz 5
March 28, 2012
Name_____________________
(please print)
Significant figures and correct units count, show charges as appropriate, and please box your answers!
1. (5 points) SHOW ALL WORK. Tin has two stable forms; white tin is the more stable phase
at temperatures higher than 13.2 °C, while the powdery gray tin is more stable at temperatures
below 13.2 °C. This phenomenon plagued Napoleon’s army in his ill-fated invasion of Russia in
1812: the tin buttons on his soldiers’ uniforms disintegrated during the Russian winter, adversely
affecting the soldiers’ health (and morale). The conversion of white to gray tin is exothermic,
with H = –2.1 kJ/mol at 13.2 °C.
a) What is S for the conversion between white and gray tin?
b) Which is the more highly ordered form of tin; white or gray?
[answers; a) At 13.2 °C = 286.35 K, G = 0 = H – TS
So, (-2.1) = (286.35)S
S = -7.3 J/mol K,
b) gray tin is more ordered]
2. (8 points) Write the balanced chemical equation that corresponds to each of the following
constants. Pay attention to phases, arrows, charges, and stoichiometry!
a) Ksp for K2SO4(s)
K2SO4(s) ⇌ 2 K+(aq) + SO42–(aq)
b) Hvap for CCl4
CCl4(l) ⇌ CCl4(g)
c)Kf for CaCO3(s)
mistaken problem! Gave 2 points to all for this one.
d) Gf for NH3(g)
½ N2(g) + 3/2 H2(g)  NH3(g)
3. (8 points) SHOW ALL WORK. In a saturated solution of Ag2CO3 at 25 °C, calculate a) the
molarity of each ion, and b) the mass that will dissolve in 0.200 L of water. (FW Ag2CO3 =
275.8 g/mol, Ksp = 8.46 x 10–12) (assume no volume change on dissolution)
Answer:
Ag2CO3(s) ⇌ 2 Ag+(aq) + CO32–(aq)
a) Ksp = [Ag+]2[CO32–] = (2x)2(x) = 8.46 x 10–12 = (4x2)(x) = 4x3
x3 = 2.115 x 10–12, so x = 1.28 x 10–4 M
[Ag+] = 2.57 x 10–4 M
[PO43–] = 1.28 x 10–4 M
b) 0.200 L H2O x (1.28 x 10–4 mol salt)/(1 L soln) x (275.8 g salt)/(1 mol salt) = 7.06 x 10–3 g
Selected Thermodynamic Data for Materials at 25 °C.
Substance
H°f (kJ/mol)
S° (J/mol K)
G°f (kJ/mol)
CaCO3(s)
CaO(s)
CCl4(l)
CO2(g)
CO(g)
Cu(s)
Cu2O(s)
H2(g)
HCl(g)
H2O(l)
H2O(g)
H2O2(l)
H2SO4(l)
KCl(s)
K2SO4(s)
N2(g)
NH3(g)
NO(g)
NO2(g)
-1207.6
-634.9
-95.7
-393.5
-110.5
0
-168.6
0
-92.3
-285.8
-241.8
-187.8
-814
-436.5
-1437.8
0
-45.9
91.3
33.2
91.7
38.1
309.9
213.8
197.7
33.2
93.1
130.7
186.9
70.0
188.7
109.6
156.9
82.6
175.6
191.6
192.8
210.8
240.1
-1129.1
-603.3
-53.6
-394.4
-137.2
0
-146.0
0
-95.3
-237.1
-228.6
-120.4
-690.0
-408.5
-1321.4
0
-16.4
87.6
51.3
O2(g)
SO3(g)
0
-395.7
205.2
256.8
0
-371.1
4. (9 points) SHOW ALL WORK. For each of the following reactions, calculate the appropriate
quantity, using tables in the back.
a. S°, in J/mol K –226.3 J/mol K
H2(g) + O2(g)  H2O2(l)
S° = [109.6] – [130.7 + 205.2] = (190.6) – (335.9) = –226.3 J/mol K
b. G°, in kJ/mol 146.0 kJ/mol,
and the thermodynamic equilibrium constant Kp, at 25 °C 3 x 10–26
Cu2O(s) ⇌ 2 Cu(s) + ½O2(g)
G = [2(0) + ½(0)] – [(–146.0)] = 146.0 kJ/mol
G = –RTln K = –(8.314 x 10–3)(298)ln K
ln K = –58.928
K = 2.56 x 10–26 but only one sig fig so 3 x 10–26