If the enthalpy of hydrogenation of $C_{6} H_{6} (l)$ into $C_{6} H_{12} (l)$ is - 205 kJ and the resonance energy of $C_{6} H_{6} (l)$ is -152 kJ/mol then the enthalpy of hydrogenation of cyclohexene is: (Assume $Δ H_{v a p}$ of $C_{6} H_{6} (l), C_{6} H_{8} (l), C_{6} H_{12} (l)$ all are equal).

- 535.5 kJ/mol

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- 238 kJ/mol

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- 357 kJ/mol

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- 119 kJ/mol

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Solution

The correct option is D $- 119 kJ/mol$
Point to remember:
1. Resonance energy $= Δ H_{e x p e r i m e n t a l} - Δ H_{c a l c u l a t e d}$

To find heat of hydrogenation of
$C_{6} H_{10} + H_{2} \to C_{6} H_{12} Δ H = ?$
We need to find the energy required to break one $C = C$ and the energy released when 2 $C - H$ bonds are formed. In benzene, 3 $C = C$ bonds break and 6 $C - H$ bonds are formed.
So $E_{b r e a k (C = C) + f o r m 2 (C = H)} = \frac{c a l c u l a t e d e n e r g y o f C_{6} H_{6}}{3}$
$E_{b r e a k (C = C) + f o r m 2 (C = H)} = \frac{- 205 + (- 152)}{3} = - 119 k J$

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C_{6} H_{6} (l)

C_{6} H_{12} (l)

C_{6} H_{6} (l)

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C_{6} H_{6} (l)

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