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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^-¹ respectively.

ΔH = ΔH_C=C + ΔH_H–H + 4ΔH_C–H – (ΔH_C–C + 6ΔH_C–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

| || |

3C(g) + 6H(g) + O(g) ———> H — C — C — C — H ΔH = ?

| |

H H

ΔH_f = [3ΔH_H–H + 1/2 ΔH_0–0 + 3ΔH_C(s)—>C(g)] – [2ΔH_C–C + 6ΔH_C–H + ΔH_C=O]

by putting the value of different bond energies you can determine the ΔH_f .

(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.

ΔH_d = 3ΔH_C–C + 3ΔH_C=C + 6ΔH_C–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) (ΔH_f⁰)C₆H₁₂ = – 156 kJ mole^–1

(iv) (ΔH⁰_f)H₂O = – 285.8 kJ mole^–1

(v) (ΔH_f⁰)CO₂ = – 393.5 kJ mole^–1

To read more, Buy study materials of Thermodynamics comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.