If for the reaction CCl_{4(g)} rightarrow C_{ | vedclass

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(B) The reaction is $CCl_{4(g)} \rightarrow C_{(g)} + 4Cl_{(g)}$.First,we find the enthalpy of formation of $CCl_{4(g)}$ using the enthalpy of vaporiz

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(B) The reaction is $CCl_{4(g)} ightarrow C_{(g)} + 4Cl_{(g)}$.First,we find the enthalpy of formation of $CCl_{4(g)}$ using the enthalpy of vaporization:$\Delta_{f} H^{	heta} (CCl_{4(g)}) = \Delta_{f} H^{	heta} (CCl_{4(l)}) + \Delta_{vap} H^{	heta} (CCl_{4(l)}) = -136.0 + 30.0 = -106.0 \ kJ \ mol^{-1}$.The enthalpy change for the reaction is given by:$\Delta_{r} H^{	heta} = \Delta_{a} H^{	heta} (C) + 4 	imes \Delta_{a} H^{	heta} (Cl) - \Delta_{f} H^{	heta} (CCl_{4(g)})$.Note that $\Delta_{a} H^{	heta} (Cl) = \frac{1}{2} \Delta_{a} H^{	heta} (Cl_2) = \frac{242.0}{2} = 121.0 \ kJ \ mol^{-1}$.$\Delta_{r} H^{	heta} = 714.0 + 4(121.0) - (-106.0) = 714.0 + 484.0 + 106.0 = 1304.0 \ kJ \ mol^{-1}$.The mean bond enthalpy of $C-Cl$ is $\frac{\Delta_{r} H^{	heta}}{4} = \frac{1304.0}{4} = 326 \ kJ \ mol^{-1}$.

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

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