At 298 K the heat of combustion of H2(g) to H2O (g) is –249 kJ mol–1. The bond energies of H–H and O=O respectively are 433 kJ mol–1 and 492 kJ mol–1. The bond energy of O–H would be :-
1. 464 kJ mol–1
2. –464 kJ mol–1
3. 232 kJ mol–1
4. –232 kJ mol–1
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11th
Chemistry
Thermodynamics
Enthalpies for Different Types of Reactions
The enthalpy of combustion ...
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Asked on December 30, 2019 by
Dola Aman
The enthalpy of combustion of H
2(g)
to give H
2
O
(g)
is −249 kJ mol
−1
and bond enthalpies of H−H and O=O are 433 kJ mol
−1
and 492 kJ mol
−1
respectively. The bond enthalpy of O−H is:
HARD
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VIDEO EXPLANATION
ANSWER
H
2
(g)+
2
1
O
2
(g)→H
2
O(g)
ΔH=−249KJ
ΔH= B.E. of H
2
+
2
1
B.E. of O
2
− B.E. of H
2
O
where B.E. is bond enthalpy
−249 =433+
2
1
×492−2×B.E.(O−H)
−249=433+246−2×B.E.(O−H)
249+433+246=2×B.E.(O−H)
B.E.(O−H)=
2
249+433+246
kJ mol
−1
B.E(O−H)=464 kJ mol
−1
Hence, the correct option is A
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