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(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-1 respectively.
ΔH = ΔHC=C + ΔHH–H + 4ΔHC–H – (ΔHC–C + 6ΔHC–H)
= (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
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ΔHf = [3ΔHH–H + 1/2 ΔH0–0 + 3ΔHC(s)—>C(g)] – [2ΔHC–C + 6ΔHC–H + ΔHC=O]
by putting the value of different bond energies you can determine the ΔHf .
(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.
C6H6 (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.
ΔHd = 3ΔHC–C + 3ΔHC=C + 6ΔHC–H
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–1
(ii) Enthalpy of hydrogenation of cyclohexene (l) = – 119 kJ mole–1
(iii) (ΔHf0)C6H12 = – 156 kJ mole–1
(iv) (ΔH0f)H2O = – 285.8 kJ mole–1
(v) (ΔHf0)CO2 = – 393.5 kJ mole–1
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