InterviewSolution
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InterviewSolution
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Calculate the resonance enegry of toulene (use Kekule structure form the following data `C_(7)H_(8)(l) +9O_(2)(g) rarr 7CO_(2)(g) +4H_(2)O(l)+ DeltaH, DeltaH^(Theta) =- 3910 kJ mol^(-1)` `C_(7)H_(8)(l) rarr C_(7)H_(8)(g), DeltaH^(Theta) = 38.1 kJ mol^(-1)` `Delta_(f)H^(Theta) (water) =- 285.8 kJ mol^(-1)` `Delta_(f)H^(Theta) [CO_(2)(g)] =- 393.5 kJ mol^(-1)` Heat of atomisaiton of `H_(2)(g) = 436.0 kJ mol^(-1)` Heat of sulimation of `C(g) = 715.0 kJ mol^(-1)` Bond energies of `C-H, C-C`, and `C=C` are `413.0, 345.6`, and `610.0 kJ mol^(-1)`. |
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Answer» From the sata given, we have `C_(7)H_(8)(l) +9O_(2)(g) rarr 7CO_(2)(g) +4H_(2)O(l) +DeltaH(-3910kJ)` It is known that `DeltaH^(Theta) ("reaction") = (sum_(f)H^(Theta))("products") -(sum_(f)H^(Theta)) ("reactants")` `= (4 xx Delta_(f)H_(H_(2)O)(l)+ 7 xx Delta_(f)H_(CO_(2))^(Theta))-Delta_(f)H[C_(7)H_(8)(l)]` or `-3910 = [4 xx (-285) +7(-393.5)] - Delta_(f)H [C_(7)H_(8)(l)]` or `Delta_(f)H^(Theta) C_(7)H_(8)(l) = (3910 - 2754.5 - 1143.2) kJ` `= 12.3 kJ mol^(-1)` `:. Delta_(f)H^(Theta)` for gaseous toluene `= (12.3 +38.1) kJ = 50.4 kJ` Chemical equation for the formation of gaseous toulene `7C(s) +4H_(2)(g) rarr C_(7)H_(8)(g)` `(Delta_(f)H^(Theta))_(cal) =[ 7 xx ("heat of sublimation of solid" C + 4BE of H_(2)(g))` `=-3 xx C=C +4 xx C-C +8 xx C-H]` `=(7 xx 715 +4 xx 436) -(3 xx 610 +4 xx 345.6 + 8xx 413)` `= (5005 +1744) - (1830 +3304 +1382.4)` `= (6749.0 - 6516.4) kJ = 232.6 kJ` Reasonance enegry `= (232.6 - 50.40) kJ = 182.2 kJ mol^(-1)` |
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