Answer
$ \Delta H^{\circ}_f{CO_{(g)}} = [ - 393.5] – [- 283.0 + 0 ] = - 110.5 kJ/mol$
Work Step by Step
$CO_{(g)} + (1/2) O_{2(g)} → CO_{2(g)}$
$\Delta H^{\circ}_{r} = -283.0 kJ/mol$.
Standard enthalpy of reaction = (sum of standard enthalpies of formation of products) – (sum of standard enthalpies of formation of reactants).
$\Delta H^{\circ}_{r} = ∑H^{\circ}_f{products} - ∑ H^{\circ}_f{reactants}$.
$\Delta H^{\circ}_{r} = [\Delta H^{\circ}_f{CO_{2(g)}} ] – [\Delta H^{\circ}_f{CO_{(g)}} + (1/2)\ times\Delta H^{\circ}_f{ O_{2(g)}}]$
Therefore the enthalpy of formation of $CO_{(g)}$ is,
$ \Delta H^{\circ}_f{CO_{(g)}} = [\Delta H^{\circ}_f{CO_{2(g)}}] - [ \Delta H^{\circ}_{r} + + (1/2)\ times\Delta H^{\circ}_f{ O_{2(g)}}]$
$ \Delta H^{\circ}_{f}CO_{2(g)} = - 393.5 kJ /mol$
$ \Delta H^{\circ}_{f}O_{2(g)} = 0 kJ /mol$
$ \Delta H^{\circ}_f{CO_{(g)}} = [ - 393.5] – [- 283.0 + 0 ] = - 110.5 kJ/mol$
