The standard enthalpies of formation of 1,3-b | vedclass

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(A) The combustion reaction for $1,3-butadiene$ is: $C_4H_{6(g)} + \frac{11}{2} O_{2(g)} 	o 4 CO_{2(g)} + 3 H_2O_{(l)}$.First,calculate the enthalpy

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$-650$

Vedclass · Answer

(A) The combustion reaction for $1,3-butadiene$ is: $C_4H_{6(g)} + \frac{11}{2} O_{2(g)} 	o 4 CO_{2(g)} + 3 H_2O_{(l)}$.First,calculate the enthalpy of combustion using standard enthalpies of formation: $\Delta H_c = [4 	imes \Delta H_f(CO_2) + 3 	imes \Delta H_f(H_2O_{(l)})] - \Delta H_f(C_4H_6)$.Given $\Delta H_f(H_2O_{(g)}) = -68 \ kcal/mol$ and $\Delta H_{vap}(H_2O) = 10 \ kcal/mol$,then $\Delta H_f(H_2O_{(l)}) = -68 - 10 = -78 \ kcal/mol$.Now,account for resonance energy: $\Delta H_{comb} = \Delta H_{comb(calc)} + (	ext{Resonance Energy of products} - 	ext{Resonance Energy of reactants})$.$\Delta H_{comb(calc)} = [4 	imes (-94) + 3 	imes (-78)] - (-30) = -376 - 234 + 30 = -580 \ kcal/mol$.Resonance correction: $\Delta H_{final} = -580 + (4 	imes (-20) - (-10)) = -580 - 80 + 10 = -650 \ kcal/mol$.

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