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Ragini Ehsaan Banerjee
Asked: 2022-11-08T17:37:10+05:30 In:

Calculate the resonance energy of gaseous benzene form the following data.
`BE(C-H) = 416.3 kJ mol^(-1)`
`BE(C-C) = 331.4 kJ mol^(-1)`
`BE(C=C) = 591.1 kJ mol^(-1)`
`Delta_(“sub”)H^(Theta)(C,”graphite”) = 718.4 kJ mol^(-1)`
`Delta_(“diss”)H^(Theta)(H_(2),g) = 435.9 kJ mol^(-1)`
`Delta_(f)H^(Theta) (“benzene”, g) = 82.9 kJ mol^(-1)`

Calculate the resonance energy of gaseous benzene form the following data.
`BE(C-H) = 416.3 kJ mol^(-1)`
`BE(C-C) = 331.4 kJ mol^(-1)`
`BE(C=C) = 591.1 kJ mol^(-1)`
`Delta_(“sub”)H^(Theta)(C,”graphite”) = 718.4 kJ mol^(-1)`
`Delta_(“diss”)H^(Theta)(H_(2),g) = 435.9 kJ mol^(-1)`
`Delta_(f)H^(Theta) (“benzene”, g) = 82.9 kJ mol^(-1)`
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  1. To compute resonance enegry, we compare the calaculated value of `Delta_(f)H^(Theta)` (benzene, g) with the given one. To calculate `Delta_(f)H^(Theta)` (benzene, g) we add the following reactions.
    (a). `6C(g) +6H(g) rarr C_(6)H_(6)(g), DeltaH^(Theta) =- (3BE_(C-C) +3BE_(C=C) +6BE_(C-H))`
    (b) `6C (“graphite”) rarr 6C(g) DeltaH^(Theta) = 6 xx 718.4 kJ mol^(-1)`
    (c) `3H_(2)(g) rarr 6H(g) DeltaH^(Theta) = 3 xx 435.9 kJ mol^(-1)`
    `ulbar(6C(“graphite”)+3H_(2)(g)rarr C_(6)H_(6)(g))`
    The corresponding enthalpy change is
    `Delta_(f)H^(Theta) =- (3BE_(C-C)+ 3BE_(C=C) +6BE_(C-H)) +[6 xx 718.4 +3 xx 435.9] kJ mol^(-1)`
    `=[-(3xx331.4 xx 591.1 +6 xx 718.4 +3 xx 435.9] kJ mol^(-1)`
    The given `Delta_(f)H^(Theta)` is `Delta_(f)H^(Theta)` (benzene, g) `= 82.9 kJ mol^(-1)`
    This means benzene become more stable by `(352.8 -82.9) kJ mol^(-1)`, i.e., `269.7 kJ mol^(-1)`
    This is its resonance enegry.

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