Given at 298 K Δ f H °( cyclohexane )=-156.0 kJ mol - Δ f H °( benzene )=+49.0 kJ mol -1 Cyclohexene + H 2 longrightarrow Cyclohexane Δ r H °=-119 kJ mol -1 The resonance energy of benzene is

Question

Given at 298 K Δ f H °( cyclohexane )=-156.0 kJ mol - Δ f H °( benzene )=+49.0 kJ mol -1 Cyclohexene + H 2 longrightarrow Cyclohexane Δ r H °=-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

Tardigrade Admin · Accepted Answer

Δr H°=Δf H°( cyclohexane )-Δf H°( benzene )= -156.0-49.0=-205 kJ

Δr H°=-119 k J In cyclohexene (C=C)= one In benzene (C=C)= three Thus, theoretical value of heat of hydrogenation of benzene =-119 × 3=-357 kJ Thus, resonance energy =-357-(-205) =-357+205=-152 kJ mol -1

Q. Given at $298 K$
$Δ_{f} H^{\circ} ($ cyclohexane $) = - 156.0 k J m o l^{-}$
$Δ_{f} H^{\circ} ($ benzene $) = + 49.0 k J m o l^{- 1}$
Cyclohexene $+ H_{2} ⟶$ Cyclohexane
$Δ_{r} H^{\circ} = - 119 k J m o l^{- 1}$
The resonance energy of benzene is

$- 152.0 k J m o l^{- 1}$

$+ 152.0 k J m o l^{- 1}$

$- 107.0 k J m o l^{- 1}$

$- 226.0 k J m o l^{- 1}$

Q. Given at 298K Δf​H∘( cyclohexane )=−156.0kJmol− Δf​H∘( benzene )=+49.0kJmol−1 Cyclohexene +H2​⟶ Cyclohexane Δr​H∘=−119kJmol−1 The resonance energy of benzene is

−152.0kJmol−1

+152.0kJmol−1

−107.0kJmol−1

−226.0kJmol−1

Q. Given at $298 K$
$Δ_{f} H^{\circ} ($ cyclohexane $) = - 156.0 k J m o l^{-}$
$Δ_{f} H^{\circ} ($ benzene $) = + 49.0 k J m o l^{- 1}$
Cyclohexene $+ H_{2} ⟶$ Cyclohexane
$Δ_{r} H^{\circ} = - 119 k J m o l^{- 1}$
The resonance energy of benzene is

$- 152.0 k J m o l^{- 1}$

$+ 152.0 k J m o l^{- 1}$

$- 107.0 k J m o l^{- 1}$

$- 226.0 k J m o l^{- 1}$