Question

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. $-208.1 kJ mol^{-1}$

Answer 1

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.

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(\[\Delta {{H}_{cyclohexene}}\]) = $-119.5 kJ mol^{-1}$
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.