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The enthalpy of hydrogenation of cyclohexene is $-119.5 kJ mol^{-1}$. If resonance energy of benzene is $-150.4 kJ mol^{-1}$, its enthalpy of hydrogenation would be:A. $-269.9 kJ mol^{-1}$B. $-358.5 kJ mol^{-1}$C. $-508.9 kJ mol^{-1}$D. $-208.1 kJ mol^{-1}$

Last updated date: 20th Jun 2024
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Hint: There are 3 double bonds in benzene and for breaking 3 double bonds, 3 ${{H}_{2}}$ molecules are required. We should remove resonance energy from total enthalpy of the reaction.

This is the hydrogenation reaction of cyclohexene:

In this reaction for breaking one double bond one hydrogen molecule is required and in benzene 3 double bonds are there so 3 hydrogen molecules are required for breaking double bonds of benzene.
Given, Enthalpy of hydrogenation of cyclohexene($\Delta {{H}_{cyclohexene}}$) = $-119.5 kJ mol^{-1}$
Here, Enthalpy of benzene ($\Delta {{H}_{benzene}}$) = 3 x (enthalpy of hydrogenation of cyclohexene)
= 3 x ($-119.5 kJ mol^{-1}$)
= $-358.5 kJ mol^{-1}$
Given, resonance energy of benzene = $-150.4 kJ mol^{-1}$
So, for calculating actual enthalpy of hydrogenation = Enthalpy of benzene – resonance energy of benzene.

From given data:
Enthalpy of hydrogenation = (-358.5) - (-150.4)
= $-208.1 kJ mol^{-1}$
So, the answer is “D”.

Note: Don’t forget to take signs of energy or enthalpy in calculation. You should remove resonance energy from the total enthalpy of reaction. For breaking every $\pi$ bond 1 molecule of hydrogen is required.