If enthalpy change for hydrogenation of ethylene is $- 132$ kJ/mole and enthalpy of formation $1, 3$ -butadiene (g) and butane(g) are $115$ kJ and $- 140$ kJ/mole respectively then calculate resonance energy of $1, 3$ -butadiene (in kJ)?

9

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18

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4

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10

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Solution

The correct option is B $9$
Enthalpy of hydrogenation of ethylene= $- 132 k J / m o l$
$∴$ Enthalpy related to hydrogenation of one $C = C$ is $- 132 k J / m o l$ .
$∴$ Calculated enthalpy of hydrogenation of $1, 3 - b u t a d i e n e = 2 \times (- 132) = - 264 k J / m o l$
Now, resonance energy of $1, 3 - b u t a d i e n e = Δ H_{e x p e c t e d} - Δ H_{a c t u a l}$
$= - 264 - (- 115 - 140)$
$= - 9$
$∴$ Resonance energy= $9 k J / m o l$

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