Chapter 6 Thermodynamics: The First Law Worksheets Answers Page 11
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12Thermodynamics: The First Law
E xample 6.21c Calculate the average C−H bond enthalpy in methane, CH
. Use the data in Appendix 2A.
4
(g) → C(g) + 4 H(g)
∆H
° = 4∆H
Solution
CH
(C−H)
4
r
B
∆H
° = ∆H
°[C(g)] + 4∆H
°[H(g)] − ∆H
°[CH
(g)]
r
f
f
f
4
−1
−1
= 716.68 + 4(217.97) − (−74.81) kJ⋅mol
= 1663.37 kJ⋅mol
−1
∆H
(C−H) = ∆H
°/4 = 415.84 kJ⋅mol
B
r
The average C−H bond enthalpy in methane is slightly larger than the average C−H
bond enthalpy in polyatomic molecules.
Example 6.21d Calculate the average C=O bond enthalpy in carbon dioxide, CO
. The enthalpy of
2
−1
formation of O(g) is 249.17 kJ⋅mol
. Use Appendix 2A for other data.
(g) → C(g) + 2 O(g)
∆H
° = 2∆H
Solution
CO
(C=O)
2
r
B
∆H
° = ∆H
°[C(g)] + 2∆H
°[O(g)] − ∆H
°[CO
(g)]
r
f
f
f
2
−1
−1
= 716.68 + 2(249.17) − (−393.51) kJ⋅mol
= 1608.53 kJ⋅mol
−1
∆H
(C=O) = ∆H
°/2 = 804.27 kJ⋅mol
B
r
The C=O double bond in carbon dioxide is considerably stronger than the average C=O bond
−1
enthalpy of 743 kJ⋅mol
in polyatomic molecules.
6.22 The Variation of Reaction Enthalpy with Temperature
Example 6.22a Calculate a value for the enthalpy of vaporization of liquid water at 298.15 K. Use
Kirchhoff ’s law to estimate the enthalpy of vaporization of liquid water at the normal
boiling point of 373.15 K. Compare the value to the one in Table 6.3 in the text. Use the
data in Appendix 2A.
O(l) → H
∆H
° = ∆H
°
Solution
H
O(g)
2
2
r
vap
∆H
° = ∆H
°[H
O(g)] − ∆H
°[H
O(l)]
r
f
2
f
2
−1
−1
= −241.82 − (−285.83) kJ⋅mol
= +44.01 kJ⋅mol
= ∆H
° at 298.15 K
vap
−1
−1
∆C
O(g)] − C
O(l)] = 33.58 − 75.29 J⋅K
⋅mol
= C
[H
[H
P
P,m
2
P,m
2
−1
−1
−1
−1
= −41.71 J⋅K
⋅mol
= −0.041 71 kJ⋅K
⋅mol
∆H
° (373.15 K) = ∆H
°(298.15 K) + ∆C
(373.15 − 298.15 K)
vap
vap
P
−1
= 44.01 + (−0.041 71)(75.00) = 44.01 − 3.13 kJ⋅mol
−1
= 40.88 kJ⋅mol
−1
The value in Table 6.3 in the text is 40.7 kJ⋅mol
. The approximation is an excellent one
in this case.
−1
Example 6.22b The bond enthalpy of H−H is 436 kJ⋅mol
at 298.15 K. Use Kirchhoff ’s law to estimate the
bond enthalpy at 0 K. Use the data in Appendix 2A.
(g) → 2 H(g)
−1
∆H
° = ∆H
Solution
H
(H−H) = 436 kJ⋅mol
at 298.15 K
2
r
B
−1
−1
∆C
[H(g)] − C
(g)] = 2(20.78) − 28.82 J⋅K
⋅mol
= 2C
[H
P
P,m
P,m
2
−1
−1
−1
−1
= +12.74 J⋅K
⋅mol
= +0.012 74 kJ⋅K
⋅mol
∆H
(0 K) = ∆H
(298.15 K) + ∆C
(0 − 298.15 K)
B
B
P
−1
= 436 + (0.012 74)(−298.15) = 436 − 3.8 kJ⋅mol
−1
= 432 kJ⋅mol
−1
The spectroscopic dissociation energy of H
(g) is 432.07 kJ⋅mol
, which corresponds to a
2
thermodynamic temperature of absolute zero. Any physical change of state that occurs at
low temperature is deliberately ignored.
11
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