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Question: The enthalpy of hydrogenation of cyclohexene is \(-119.5 kJ mol^{-1}\). If resonance energy of benze...

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

Explanation

Solution

Hint: There are 3 double bonds in benzene and for breaking 3 double bonds, 3 H2{{H}_{2}} molecules are required. We should remove resonance energy from total enthalpy of the reaction.

Complete step-by-step answer:
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(ΔHcyclohexene\Delta {{H}_{cyclohexene}}) = −119.5kJmol−1-119.5 kJ mol^{-1}
Here, Enthalpy of benzene (ΔHbenzene\Delta {{H}_{benzene}}) = 3 x (enthalpy of hydrogenation of cyclohexene)
= 3 x (−119.5kJmol−1-119.5 kJ mol^{-1})
= −358.5kJmol−1-358.5 kJ mol^{-1}
Given, resonance energy of benzene = −150.4kJmol−1-150.4 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)
= −208.1kJmol−1-208.1 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 π\pi bond 1 molecule of hydrogen is required.