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Chapter 18: Thermodynamics: Directionality of Chemical Reactions
852
89
.
Answer:
Reaction (b) needs 5.068 g; reactions (c) needs 26.94 g.
Strategy and Explanation:
Adapt the method described in the solution to Question 6.85. Find the Gibbs free
energy of the graphite reaction and use that as a thermochemical conversion factor to supply the Gibbs free
energy of the reactions that require work be done in Question 87 (b) and (c).
C(graphite) + O
2
(g)
CO
2
(g)
This equation is identical to the equation describing the formation reaction for CO
2
, so
Δ
G° = –394.359 kJ/mol
The calculation for 87(b) looks like this:
166.4 kJ endergonic reaction
"
1 mol C
394.359 kJ provided
"
12.011 g C
1 mol C
=
5.068 g C
The rest of the results are described in the table below.
Endergonic Reaction
Reference
Calculated
Δ
G° (X)
(kJ)
Mass of graphite needed
(g)
87(b)
166.4 kJ
5.068
87(c)
884.5 kJ
26.94
90.
Answer:
Reaction (a) needs 13.945 g; reactions (c) needs 0.89046 g.
Strategy and Explanation:
Adapt the method described in the solution to Question 6.85. Find the Gibbs free
energy of the hydrogen reaction and use that as a thermochemical conversion factor to supply the Gibbs free
energy of the reactions that require work be done in Question 88 (a) and (c).
H
2
(g) + O
2
(g)
H
2
O(g)
This equation is identical to the formation reaction equation for H
2
O(g), so
Δ
G° = –228.572 kJ/mol.
The calculation for 88(a) looks like this:
1582.3 kJ endergonic reaction
"
1 mol H
2
228.572 kJ provided
"
2.0158 g H
2
1 mol H
2
=
13.954 g H
2
The rest of the results are described in the table below.
Endergonic Reaction
Reference
Calculated
Δ
G° (X)
(kJ)
Mass of H
2
needed
(g)
88(a)
1582.3 kJ
13.945
88(c)
100.97 kJ
0.89046
91
.
Answer:
(a) 2 C(s) + 2 Cl
2
(g) + TiO
2
(s)
TiCl
4
(g) + 2 CO(g) (b)
Δ
H° = –44.6 kJ;
Δ
S° = 242.7 J K
–1
;
Δ
G° = –116.5 kJ (c) productfavored (d) more reactantfavored
Strategy and Explanation:
Balance the equation, then use the method described in the solution to Questions 51
and 57.
(a)
2 C(s) + 2 Cl
2
(g) + TiO
2
(s)
TiCl
4
(g) + 2 CO(g)
Check: 2 C, 4 Cl, 1 Ti, 2 O
(b)
Δ
X° = (1 mol)
×
X°(TiCl
4
) + (2 mol)
×
X°(CO) – (2 mol)
×
X°(C)
– (2 mol)
×
X°(Cl
2
) – (1 mol)
×
X°(TiO
2
) (X =
Δ
H
f
, S, or
Δ
G
f
)
Δ
H° = (1 mol)
×
(–763.2 kJ/mol) + (2 mol)
×
(–110.525 kJ/mol)
– (2 mol)
×
(0 kJ/mol) – (2 mol)
×
(0 kJ/mol)
– (1 mol)
×
(–939.7 kJ/mol) = –44.6 kJ
Δ
S° = (1 mol)
×
(354.9 J K
–1
mol
–1
) + (2 mol)
×
(197.674 J K
–1
mol
–1
)
– (2 mol)
×
(5.740 J K
–1
mol
–1
) – (2 mol)
×
(223.066 K
–1
mol
–1
)
– (1 mol)
×
(49.92 J K
–1
mol
–1
) = 242.7 J K
–1
Δ
G° = (1 mol)
×
(–726.7 kJ/mol) + (2 mol)
×
(–137.168 kJ/mol)
– (2 mol)
×
(0 kJ/mol) – (2 mol)
×
(0 kJ/mol)
– (1 mol)
×
(–884.5 kJ/mol) = –116.5 kJ
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View Full DocumentChapter 18: Thermodynamics: Directionality of Chemical Reactions
853
(c)
Δ
G° is negative, so the reaction is productfavored.
(d) As the reaction temperature increases, heat energy is added to the system. The reaction is
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 Fall '08
 Kerber
 Chemistry, Reaction

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