If enthalpies of formation of C_2H_{4(g)},CO_ | vedclass

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(B) The combustion reaction for $C_2H_{4(g)}$ is: $C_2H_{4(g)} + 3O_{2(g)} 	o 2CO_{2(g)} + 2H_2O_{(l)}$$\Delta H_{combustion} = \sum \Delta H_f^o(	e

Vedclass · Accepted Answer

$-1412$

Vedclass · Answer

(B) The combustion reaction for $C_2H_{4(g)}$ is: $C_2H_{4(g)} + 3O_{2(g)} 	o 2CO_{2(g)} + 2H_2O_{(l)}$$\Delta H_{combustion} = \sum \Delta H_f^o(	ext{products}) - \sum \Delta H_f^o(	ext{reactants})$$\Delta H_{combustion} = [2 	imes \Delta H_f^o(CO_2) + 2 	imes \Delta H_f^o(H_2O)] - [\Delta H_f^o(C_2H_4) + 3 	imes \Delta H_f^o(O_2)]$Given: $\Delta H_f^o(C_2H_4) = 52 \ kJ \ mol^{-1}$,$\Delta H_f^o(CO_2) = -394 \ kJ \ mol^{-1}$,$\Delta H_f^o(H_2O) = -286 \ kJ \ mol^{-1}$,and $\Delta H_f^o(O_2) = 0 \ kJ \ mol^{-1}$.$\Delta H_{combustion} = [2(-394) + 2(-286)] - [52 + 3(0)]$$\Delta H_{combustion} = [-788 - 572] - 52 = -1360 - 52 = -1412 \ kJ \ mol^{-1}$.

Bond	Bond energy $(kJ \ mol^{-1})$
$C-H$	$413$
$C-C$	$348$
$C=C$	$610$
$H-H$	$436$

If enthalpies of formation of $C_2H_{4(g)}$,$CO_{2(g)}$ and $H_2O_{(l)}$ at $25 \ ^\circ C$ and $1 \ atm$ pressure are $52$,$-394$ and $-286 \ kJ \ mol^{-1}$ respectively,the enthalpy of combustion of $C_2H_{4(g)}$ will be.....$kJ \ mol^{-1}$.

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$C-C$ $348$

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