Chemical Thermodynamics Question 466

Question: If (i) $ \Delta H_{f}^{o} $ (benzene) $ =-358.5kJmo{{l}^{-1}} $ .

(ii) Heat of atomization of graphite $ =716.8kJ,mo{{l}^{-1}} $ . (iii) Bond energy of $ C-H,C-C,C=C $ and $ H-H $ bonds are 490, 340, 620 and $ 436.9kJmo{{l}^{-1}} $ respectively. The resonance energy (in $ kJmo{{l}^{-1}} $ ) of $ C_6H_6 $ using Kekule formula is

Options:

A) - 150

B) - 50

C) - 250

D) +150

Show Answer

Answer:

Correct Answer: A

Solution:

  • $ 6C(g)+3H_2(g)\xrightarrow{,},C_6H_6; $

$ \Delta {H_{\exp }}=-358,kJ $

$ \Delta H_{f} $ can also be calculated as; $ \Delta H_{f}(C_6H_6)=[6\times \Delta H_{C}(s)\xrightarrow{{}} $

$ {C_{(g)}}+3\times \Delta {H_{H-H}}] $

$ -[3BE $ of $ C=C+3BE $ of $ C=C,+6BE $ of $ C-H] $

$ =[6\times 716.8+3\times 436.9] $

$ -[3\times 340+3\times 620+6\times 490] $

$ \Delta H_{resonance}=\Delta H_{f}{(\exp )}-\Delta {H{f(calc)}} $

$ =-358.5-(-208.5)=-150,kJ,mo{{l}^{-1}} $



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