Answer to Question #215862 in Physical Chemistry for Yougankshi

Question

Using the following data, calculate the amount of energy by which benzene is more

stable as compared to one Kekule structure:



∆fH*=(methane, g) = −84.85 kJ/mol

∆fH*=(ethane, g) = −94.85 kJ/mol

∆fH*=(ethylene, g) = −65.35 kJ/mol

∆fH*=(benzene, g) = −98.45 kJ/mol



C (graphite) → C (g); ∆H* = 714.39 kJ/mol



H2(g) → 2H(g) ∆H* = 435.89 kJ/mol



(here '*' denotes degree sign)

Answer 1

For C6

​H6​; 6C(s)+3H2​(g)⟶C6​H6

​; ;ΔHexp

​=−358.5kJ

H2(g) → 2H(g) ∆H* = 435.89 kJ/mol

Also, using Bond energy ΔHcal

​ can be given as ΔHcal

​−Bondenergydatausedforformationofbond+B.E.datausedfordissociationofbonds=−[3×e(C−−C)​+3×e(C==C)​+6×e(C−−H)​]+[6Cs→g​+3×e(H−−H)​]

=−[3×340+3×620+6×490] +[6×716.8+3×436.9]ΔHcal

​=−208.5kJ + 435 -93 + 714-94.85-84.85-65.35-98.45

=

∴Resonanceenergy=ΔHexp​−ΔHcal

​

                                     =−358.5−(−319.85)=−38.65kJ