Chemical Thermodynamics Question 344
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}} $
BETA
