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Hess' Law

Calculate the...

Question

Calculate the resonance energy of $C_{6} H_{6}$ , using Kekule formula for $C_{6} H_{6}$ from the following data: (i) $Δ H^{\circ}$ for $C_{6} H_{6} = - 358.5 k J m o l^{- 1}$ (ii) $H e a t o f a t o m i s a t i o n o f C = 716.8 k J m o l^{- 1}$ (iii) Bond energy of $C - - H, C - - C, C = = C$ and $H - - H$ are 490, 340, 620 and $436.9 k J m o l^{- 1}$ . Report your answer in kj (only value).

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Solution

For $C_{6} H_{6}$ ; $6 C (s) + 3 H_{2} (g) ⟶ C_{6} H_{6};; Δ H_{e x p} = - 358.5 k J$
Also, using Bond energy $Δ H_{c a l}$ can be given as $Δ H_{c a l} - B o n d e n e r g y d a t a u s e d f o r f o r m a t i o n o f b o n d + B . E . d a t a u s e d f o r d i s s o c i a t i o n o f b o n d s$
$= - [3 \times e_{(C - - C)} + 3 \times e_{(C = = C)} + 6 \times e_{(C - - H)}] + [6 C_{s \to g} + 3 \times e_{(H - - H)}]$
$= - [3 \times 340 + 3 \times 620 + 6 \times 490] + [6 \times 716.8 + 3 \times 436.9] Δ H_{c a l} = - 208.5 k J$
$∴ R e s o n a n c e e n e r g y = Δ H_{e x p} - Δ H_{c a l}$
$= - 358.5 - (- 208.5) = - 150.0 k J$

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