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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 238 kJ/mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 357 kJ/mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 119 kJ/mol

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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 $E_{b r e a k (C = C) + f o r m 2 (C = H)} = \frac{c a l c u l a t e d e n e r g y o f C_{6} H_{6}}{3}$
$E_{b r e a k (C = C) + f o r m 2 (C = H)} = \frac{- 205 + (- 152)}{3} = - 119 k J$

Suggest Corrections

Similar questions

C_{6} H_{6} (l)

C_{6} H_{12} (l)

C_{6} H_{6} (l)

Δ H_{v a p}

C_{6} H_{6} (l), C_{6} H_{8} (l), C_{6} H_{12} (l)

- 119.5 kJ/ mol

- 150.4 kJ/mol

- 119.5 k J / m o l

- 150.4 k J / m o l

- 119.5 k J / m o l

- 150.4 k J / m o l

Join BYJU'S Learning Program