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Home » Study Material » IIT JEE Physics » Mechanics » Thermal Physics »
Application of Bond Energies
(i) Determination of enthalpies of reactions
Suppose we want to determine the enthalpy of the
reaction.
If bond energies given for C ¾ C, C = C, C¾H, and H ¾ H
are 347.3, 615.0, 416.2 and 435.1KJ mol respectively.
ΔH = ΔH + ΔH + 4ΔH – (ΔH + 6ΔH )
= (615.0 + 435.1) – (347.3 + 832.4) => –129.6 KJ
(ii) Determination of enthalpies of formation of
compounds
Consider the formation of acetone.
H O H
| || |
3C(g) + 6H(g) + O(g) ———> H — C — C — C — H
ΔH = ?
| |
H H
ΔH = [3ΔH + 1/2 ΔH + 3ΔH ] – [2ΔH +
6ΔH + ΔH ]
by putting the value of different bond energies you can
determine the ΔH .
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Application of bond energies
-1
C=C H–H C–H C–C C–H
f H–H 0–0 C(s)—>C(g) C–C
C–H C=O
f
Name
OR
(iii) Determination of resonance energy
If a compound exhibits resonance, there is a
considerable difference between the enthalpies of
formation as calculated from bond energies and those
determined experimentally. As an example we may
consider the dissociation of benzene.
C H (g) ———> 6C(g) + 6H(g)
Assuming that benzene ring consists of three single and
three double bonds (Kekule’s structure) the calculated
dissociation energy comes out to be 5384.1 KJ from bond
energies data.
ΔH = 3ΔH + 3ΔH + 6ΔH
The experimental value is known to be 5535.1 KJ/mol.
Evidently, the energy required for the dissociation of
benzene is 151 KJ more that the calculated value. The
difference of 151 KJ gives the resonance energy of
benzene.
Exercise:
Calculate the enthapy of combustion of benzene (l) on
the basis of the following.
(i) Resonance energy of benzene (l) = – 152 kJ mole
(ii) Enthalpy of hydrogenation of cyclohexene (l) = –
119 kJ mole
(iii) (ΔH )C H = – 156 kJ mole
(iv) (ΔH )H O = – 285.8 kJ mole
(v) (ΔH )CO = – 393.5 kJ mole
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