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(A) The reaction is: $C_2H_{4(g)} + H_{2(g)} 	o C_2H_{6(g)}$Change in enthalpy $(\Delta H)$ is calculated as: $\Delta H = \sum 	ext{Bond energy of r

Vedclass · Accepted Answer

$-128$

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(A) The reaction is: $C_2H_{4(g)} + H_{2(g)} 	o C_2H_{6(g)}$Change in enthalpy $(\Delta H)$ is calculated as: $\Delta H = \sum 	ext{Bond energy of reactants} - \sum 	ext{Bond energy of products}$Reactants:$1 	imes (C=C) = 610 \ kJ$$4 	imes (C-H) = 4 	imes 413 = 1652 \ kJ$$1 	imes (H-H) = 436 \ kJ$Total reactant bond energy = $610 + 1652 + 436 = 2698 \ kJ$Products:$1 	imes (C-C) = 348 \ kJ$$6 	imes (C-H) = 6 	imes 413 = 2478 \ kJ$Total product bond energy = $348 + 2478 = 2826 \ kJ$$\Delta H = 2698 - 2826 = -128 \ kJ$Thus,the change in heat content is $-128 \ kJ$.

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

With the help of the following data,find out the change in heat content for the reaction in $kJ$:
$C_2H_{4(g)} + H_{2(g)} \to C_2H_{6(g)}$

Bond Bond energy $(kJ \ mol^{-1})$

$C-H$ $413$

$C-C$ $348$

$C=C$ $610$

$H-H$ $436$

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With the help of the following data,find out the change in heat content for the reaction in $kJ$:$C_2H_{4(g)} + H_{2(g)} \to C_2H_{6(g)}$ Bond Bond energy $(kJ \ mol^{-1})$ $C-H$ $413$ $C-C$ $348$ $C=C$ $610$ $H-H$ $436$