Chem 1B
Dr. White
Worksheet 15 Key– Crossover Temperature and ΔG°
1. The enthalpy and entropy change of a reaction are -4.7 kJ/mole and -69.5 J/mole K
respectively over a wide range of temperatures. What is the crossover temperature? Is the
reaction spontaneous at temperatures above or below this temperature?
0 = ΔH − TΔS
ΔH
−4.7kJ / mol
T=
=
= 68K
ΔS −0.0695kJ / molK
The reaction is spontaneous below 68K and non-spontaneous above 68K.
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2. The enthalpy and entropy change of a reaction are 34.7 kJ/mole and 6.95 J/mole K
respectively over a wide range of temperatures. What is the crossover temperature? Is the
reaction spontaneous at temperatures above or below this temperature?
0 = ΔH − TΔS
ΔH
34.7kJ / mol
T=
=
= 4990K
ΔS 0.00695kJ / molK
The reaction is spontaneous above 4990 K and non-spontaneous below 4990 K.
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3. Using the listed [ΔGof values] calculate ΔG° for the reaction:
Mn3O4(s) [-1283.2 kJ/mol] + 4 CO(g) [-137.2 kj/mol] → 3 Mn(s) + 4 CO2(g) [-394.4 kJ/mol]
ΔG o rxn = [4mol × −394.4kJ / mol + 3mole × 0.0kJ / mol]−
[1mol × −1283.2kJ / mol + 4mol × −137.2kJ / mol] = 254kJ
4. Using the listed information calculate ΔG° (kJ) for the reaction at 25°C and 100.°C:
C6H12(g) + 9 O2(g) → 6 CO2(g) + 6 H2O(l)
ΔHof (kJ/mole)
-156.4
-393.5
-285.9
o
S (J/K mole)
204.4
205.1
213.7
69.9
(NOTE: the ΔH° and ΔS° values are constant over 25°C and 100°C)
ΔH o rxn = [6mol × −393.5kJ / mol + 6mole × −285.9kJ / mol]−
[1mol × −156.4kJ / mol + 9mol × 0kJ / mol] = −3920kJ
ΔS o rxn = [6mol × 213.7J / molK + 6mole × 69.9J / molK ]−
[1mol × 204.4 J / molK + 9mol × 205.1kJ / mol] = −348J / K
ΔG o = −3920kJ / mol − 298K × −0.348kJ / K = −3816kJ
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ΔG o = −3920kJ / mol − 373K × −0.348kJ / K = −3790kJ
1 Chem 1B
Dr. White
2 5. Calculate ΔHof at 298K for ClF3 from the ΔG° of reaction and the S° values.
3 F2(g) + Cl2(g) → 2 ClF3(g)
ΔG° = -246.0 kJ
S° (J/K.mole)
202.8 223.1
281.6
ΔS o rxn = [2mol × 281.6J / molK ]− [3mol × 202.8J / molK + 1mol × 223.1kJ / mol] = −268.3J / K
ΔG o = ΔH o − TΔS o
−246.0kJ = ΔH o − 298K (−0.2683kJ / K )
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ΔH o = −326.0kJ
−326.0kJ = [2mol × ΔH of ,ClF3 ]− 0
ΔH of ,ClF3 = −163.0kJ / mol
6. From the table of values below, determine the boiling point of bromine (in °C).
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Br2(l)
Br2(g)
Br(g)
Br–(g)
∆H°f
[kJ/mol]
—
30.91
111.9
–218.9
∆Gf°
[kJ/mol]
—
3.13
82.40
–102.9
S°
[J/mol.K]
152.23
245.38
174.90
80.71
Br2 (l)  Br2 (g)
Boiling point occurs when ΔG°=0 (when ΔG°<0, the gas phase is favored and when ΔG°>0,
the liquid phase is favored).
ΔS o rxn = 245.38J / molK −152.23J / molK
ΔS o rxn = 93.15J / molK
ΔH o rxn = 30.91kJ / mol − 0 = 30.91kJ / mol
0 = ΔH o − TΔS o
ΔH o
30.91kJ / mol
T=
=
= 331.8304K
o
0.09315kJ / molK
ΔS
331.8304 − 273.15 = 58.7 o C
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