Question

Given at $298 \,K$
$\Delta_{ f } H ^{\circ}($ cyclohexane $)=-156.0\, kJ \,mol ^{-}$
$\Delta_{ f } H ^{\circ}($ benzene $)=+49.0\, kJ \,mol ^{-1}$
Cyclohexene $+ H _{2} \longrightarrow $ Cyclohexane
$\Delta_{ r } H ^{\circ}=-119 \,kJ \,mol ^{-1}$
The resonance energy of benzene is

• A. $-152.0 \,kJ\, mol ^{-1}$
• B. $+152.0 \,kJ \,mol ^{-1}$
• C. $-107.0 \,kJ\, mol ^{-1}$
• D. $-226.0 \,kJ \,mol ^{-1}$

Answer 1

Answer: (A)

$\Delta_{r} H^{\circ}=\Delta_{f} H^{\circ}($ cyclohexane $)-\Delta_{f} H^{\circ}($ benzene $)=$

$-156.0-49.0=-205 \,kJ$



$\Delta_{r} H^{\circ}=-119\, k J$

In cyclohexene $(C=C)=$ one

In benzene $(C=C)=$ three

Thus, theoretical value of heat of hydrogenation of benzene

$=-119 \times 3=-357 \,kJ$

Thus, resonance energy $=-357-(-205)$

$=-357+205=-152\,kJ\, mol ^{-1}$