HW 9-5: pg. 381 #69, 72, 88 (HINT: You are solving for the ΔH for the formation of CH4. The formation of
CH4 is C + 2 H2 → CH4), 89
69
From the following data, calculate the average bond energy for the N–H bond.
NH3(g) → NH2(g) + H(g)
ΔH° = 435 kJ/mol
NH2(g) → NH(g) + H(g)
ΔH° = 381 kJ/mol
NH(g) → N(g) + H(g)
ΔH° = 360 kJ/mol
NH3(g) → N(g) + 3H(g)
ΔH° = 1176 kJ/mol
NH3(g) → N(g) + 3H(g)
ΔH° = 3 DN−H
The average bond energy, DN−H =
72
1176 kJ/mol
= 392 kJ / mol
3
For the reaction
2 C2H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (g)
(a) Predict the enthalpy of the reaction from the average bond energies in table 9.4.
H
H
2
C
H
H
C
H
H
Bonds Broken
C−H
C−C
O=O
Bonds Formed
C=O
O−H
+ 7
O
O
4O
Number Broken
12
2
7
Number Formed
8
12
C
O
+
6
O
H
H
Bond Energy
(kJ/mol)
414
347
498.7
Energy Needed
(kJ)
4968 (Endo)
694 (Endo)
3491 (Endo)
Bond Energy
(kJ/mol)
799
460
Energy Released
(kJ)
6392 (Exo)
5520 (Exo)
ΔH° = total energy input − total energy released
= 4968 + 694 + 3491− (6392 + 5520) = −2759 kJ/mol
(b) Calculate the enthalpy of reaction from the standard enthalpies of formation (see Appendix
3) of the reactant and product molecules, and compare the result with your answer in part
(a)
ΔH ° = 4ΔH !f (CO2 ) + 6ΔH !f (H 2O) − [2ΔH !f (C 2 H 6 ) + 7ΔH !f (O2 )]
ΔH° = (4)(−393.5 kJ/mol) + (6)(−241.8 kJ/mol) − [(2)(−84.7 kJ/mol) + (7)(0)]
= −2855 kJ/mol
The answers for part (a) and (b) are close but different, because average bond energies are used for part
(a).
2
88
Homework #9-5 Answer key
Using the following information and the fact that the average C–H bond energy is 414 kJ/mol,
estimate the standard enthalpy of formation of methane, CH4.
→ CH4(g)
The formation of CH4 from its elements is: C(s) + 2H2(g) ⎯⎯
The reaction could take place in two steps:
→ C(g) ΔH = 716 kJ/mol
Step 1: C(s) ⎯⎯
→ 4H(g) ΔH = 872.8 kJ/mol (Endo because break bonds)
Step 2: 2 H2(g) ⎯⎯
→ CH4(g)
Step 3: C(g) + 4H(g) ⎯⎯
ΔH = – 4 DC–H = −4 × 414 kJ/mol = −1656 kJ/mol (Exo because making bonds)
Therefore, ΔH !f (CH 4 ) would be approximately the sum of the enthalpy changes for the three steps.
ΔH !f (CH 4 ) = 716 + 872.8 − 1656 kJ/mol = − 67 kJ / mol
This is close to the actual value. The ΔH !f (CH 4 ) = − 74.85 kJ/mol.
89
Based on energy considerations, which of the following reactions will occur more readily?
(Hint: Refer to table 9.4, and assume that the average bond energy of the C–Cl bond is 338
kJ/mol.)
(a) Cl (g) + CH4 (g) → CH3Cl (g) + H(g)
Bond broken:
Bond made:
C−H
C−Cl
DC–H = 414 kJ/mol
DC–Cl = −338 kJ/mol
!
ΔH rxn
= 414 − 338 = 76 kJ/mol
(b) Cl (g) + CH4 → CH3(g) + HCl (g)
Bond broken:
Bond made:
C−H
H−Cl
DC–H = 414 kJ/mol
DH–Cl = −431.9 kJ/mol
!
ΔH rxn
= 414 − 431.9 = − 18 kJ/mol
Based on energy considerations, reaction (b) will occur readily since it is exothermic. Reaction (a)
is endothermic.