Calculate the enthalpy of combustion of benzene (1) (in kJ/mol) on the basis of the following data : (i) Resonance energy of benzene(l) = -152 kJ/mol (ii) Enthalpy of hydrogenation of cyclohexene ( l

Question

Calculate the enthalpy of combustion of benzene (1) (in kJ/mol) on the basis of the following data :

(i) Resonance energy of benzene(l) = -152 kJ/mol

(ii) Enthalpy of hydrogenation of cyclohexene $(l) = - 119 kJ / mol$

(iii) $Δ H_{f}^{\circ}$ of $C_{6} H_{12} (ℓ) = - 156 kJ / mol$

(iv) $Δ H^{\circ}_{f}$ of $H_{2} O (ℓ) = - 285.8 kJ / mol$

(v) $Δ H_{f}^{\circ}$ of $CO_{2} (g) = - 393.5 kJ / mol$

TestHub · Accepted Answer

Hydrogenation of benzene:

C₆H₆(l) + 3H₂ → C₆H₁₂(l)

For hypothetical cyclohexatriene:

$Δ H = 3 (- 119) = - 357 k J$

Resonance energy = $- 152$ kJ, so benzene hydrogenation:

$Δ H = - 357 + 152 = - 205 k J$

$Δ H_{f} (C_{6} H_{6}) = Δ H_{f} (C_{6} H_{12}) - Δ H_{h y d}$

$= - 156 - (- 205) = 49 k J / m o l$

Combustion:

C₆H₆ + $\frac{15}{2}$ O₂ → 6CO₂ + 3H₂O

$Δ H_{c} = [6 (- 393.5) + 3 (- 285.8)] - 49$

$= - 3267.4 k J / m o l$

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