Enthalpies of formation of CCl_{4(g)},H_2O_{( | vedclass

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(C) The enthalpy of reaction is calculated using the formula: $\Delta_{r}H^{\circ} = \sum \Delta_{f}H^{\circ}(	ext{products}) - \sum \Delta_{f}H^{\ci

Vedclass · Accepted Answer

$173$

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(C) The enthalpy of reaction is calculated using the formula: $\Delta_{r}H^{\circ} = \sum \Delta_{f}H^{\circ}(	ext{products}) - \sum \Delta_{f}H^{\circ}(	ext{reactants})$For the reaction: $CCl_{4(g)} + 2H_2O_{(g)} ightarrow CO_{2(g)} + 4HCl_{(g)}$$\Delta_{r}H^{\circ} = [\Delta_{f}H^{\circ}(CO_{2(g)}) + 4 	imes \Delta_{f}H^{\circ}(HCl_{(g)})] - [\Delta_{f}H^{\circ}(CCl_{4(g)}) + 2 	imes \Delta_{f}H^{\circ}(H_2O_{(g)})]$Substituting the given values:$\Delta_{r}H^{\circ} = [-394 + 4 	imes (-92)] - [-105 + 2 	imes (-242)]$$\Delta_{r}H^{\circ} = [-394 - 368] - [-105 - 484]$$\Delta_{r}H^{\circ} = -762 - (-589)$$\Delta_{r}H^{\circ} = -762 + 589 = -173 \ kJ \ mol^{-1}$The magnitude of the enthalpy of the reaction is $173 \ kJ \ mol^{-1}$.

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

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