Given Delta H_f^{circ} for CO_{2(g)}, CO_{(g) | vedclass

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(D) The reaction is: $CO_{2(g)} + H_{2(g)} \longrightarrow CO_{(g)} + H_2 O_{(g)}$$\Delta H_r^{\circ} = \sum \Delta H_{f, 	ext{products}}^{\circ} - \

Vedclass · Accepted Answer

$41.2$

Vedclass · Answer

(D) The reaction is: $CO_{2(g)} + H_{2(g)} \longrightarrow CO_{(g)} + H_2 O_{(g)}$$\Delta H_r^{\circ} = \sum \Delta H_{f, 	ext{products}}^{\circ} - \sum \Delta H_{f, 	ext{reactants}}^{\circ}$$\Delta H_r^{\circ} = [\Delta H_{f, CO}^{\circ} + \Delta H_{f, H_2 O}^{\circ}] - [\Delta H_{f, CO_2}^{\circ} + \Delta H_{f, H_2}^{\circ}]$Since the standard enthalpy of formation for an element in its standard state is zero, $\Delta H_{f, H_2}^{\circ} = 0$.$\Delta H_r^{\circ} = [-110.5 + (-241.8)] - [-393.5 + 0]$$\Delta H_r^{\circ} = -352.3 + 393.5 = 41.2 \ kJ \ mol^{-1}$

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

Given $\Delta H_f^{\circ}$ for $CO_{2(g)}$, $CO_{(g)}$ and $H_2 O_{(g)}$ are $-393.5$, $-110.5$ and $-241.8 \ kJ \ mol^{-1}$, respectively. The $\Delta H_r^{\circ}$ [in $kJ \ mol^{-1}$] for the reaction $CO_{2(g)} + H_{2(g)} \longrightarrow CO_{(g)} + H_2 O_{(g)}$ is

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$C-C$ $348$

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