If enthalpy of hydrogenation of $C_{6} H_{6} (l)$ into $C_{6} H_{12} (l)$ is $- 205 k J$ and resonance energy of $C_{6} H_{6} (l)$ is $- 152 k J / m o l$ then enthalpy of hydrogenation of the given figure is
[Assume : $Δ H_{v a p}$ of $C_{6} H_{6} (l), C_{6} H_{8} (l), C_{6} H_{12} (l)$ are all equal]

+ 535.5 k J / m o l

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- 238 k J / m o l

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- 357 k J / m o l

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- 119 k J / m o l

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Solution

The correct option is D $- 119 k J / m o l$
Total enthalpy of hydrogenation if resonance doesn't happen $= - 205 + (- 152) = - 357 k J / m o l$
So for given molecule (single double bond) energy of hydrogenation is $- 357 / 3 = - 119 k J / m o l$

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Similar questions

Q. 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}

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

all are equal).

Q. The standard enthalpies of combustion of

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

C_{(g r a p h i t e)}

and

H_{2} (g)

are respectively

- 3270

k J

{m o l}^{- 1},

- 394

k J

{m o l}^{- 1}

and

- 286

k J

{m o l}^{- 1}

. What is the standard enthalpy of formation of

C_{6} H_{6} (l)

k J

{m o l}^{- 1}

The standard molar enthalpies of formation of cyclohexane (l) and benzene (l) at $25^{\circ} C$ are -156 and +49 kJ/mol. The standard enthalpy of hydrogenation of cyclohexene (l) at $25^{\circ} C$ is -119 kJ/mol. Estimate the magnitude of resonance energy.